commutator theory for congruence modular varieties ralph freese

Commutator Theory for

Congruence Modular Varieties

Ralph Freese and Ralph McKenzie

Contents

Introduction 1

Chapter 1. The Commutator in Groups and Rings 7Exercise 10

Chapter 2. Universal Algebra 11Exercises 19

Chapter 3. Several Commutators 21Exercises 22

Chapter 4. One Commutator in Modular Varieties;Its Basic Properties 25Exercises 33

Chapter 5. The Fundamental Theorem on Abelian Algebras 35Exercises 43

Chapter 6. Permutability and a Characterization ofModular Varieties 47Exercises 49

Chapter 7. The Center and Nilpotent Algebras 53Exercises 57

Chapter 8. Congruence Identities 59Exercises 68

Chapter 9. Rings Associated With Modular Varieties: Abelian Varieties 71Exercises 87

Chapter 10. Structure and Representationin Modular Varieties 891. Birkhoff-Jonsson Type Theorems For Modular Varieties 892. Subdirectly Irreducible Algebras inFinitely Generated Varieties 923. Residually Small Varieties 974. Chief Factors and Simple Algebras 102Exercises 103

Chapter 11. Joins and Products of Modular Varieties 105

Chapter 12. Strictly Simple Algebras 109

iii

iv CONTENTS

Chapter 13. Mal’cev Conditions for Lattice Equations 115Exercises 120

Chapter 14. A Finite Basis Result 121

Chapter 15. Pure Lattice Congruence Identities 1351. The Arguesian Equation 139

Related Literature 141

Solutions To The Exercises 147Chapter 1 147Chapter 2 147Chapter 4 148Chapter 5 150Chapter 6 152Chapter 7 156Chapter 8 158Chapter 9 161Chapter 10 165Chapter 13 165

Bibliography 169

Index 173

Introduction

In the theory of groups, the important concepts of Abelian group,solvable group, nilpotent group, the center of a group and centraliz-ers, are all defined from the binary operation [x, y] = x−1y−1xy. Eachof these notions, except centralizers of elements, may also be definedin terms of the commutator of normal subgroups. The commutator[M,N] (where M and N are normal subgroups of a group) is the (nor-mal) subgroup generated by all the commutators [x, y] with x ∈ M ,y ∈ N . Thus we have a binary operation in the lattice of normalsubgroups. This binary operation, in combination with the lattice op-erations, carries much of the information about how a group is puttogether. The operation is also interesting in its own right. It is a com-mutative, monotone operation, completely distributive with respect tojoins in the lattice.

There is an operation naturally defined on the lattice of ideals ofa ring, which has these properties. Namely, let [J,K] be the idealgenerated by all the products jk and kj, with j ∈ J and k ∈ K. Thecongruity between these two contexts extends to the following facts:[M,M] is the smallest normal subgroup U of M for which M/U is acommutative group; [J,J] is the smallest ideal K of J for which the ringJ/K is a commutative group; that is, a ring with trivial multiplication.

Now it develops, amazingly, that a commutator can be definedrather naturally in the congruence lattices of every congruence mod-ular variety. This operation has the same useful properties that thecommutator for groups (which is a special case of it) possesses. Theresulting theory has many general applications and, we feel, it is quitebeautiful.

In this book we present the basic theory of commutators in congru-ence modular varieties and some of its strongest applications. The bookby H. P. Gumm [41] offers a quite different approach to the subject.Gumm developed a sustained analogy between commutator theory andaffine geometry which allowed him to discover many of the basic factsabout the commutator. We take a more algebraic approach, using someof the shortcuts that Taylor and others have discovered.

1

2 INTRODUCTION

Historical remarks. The lattice of normal subgroups of a group,with the commutator operation, is a lattice ordered monoid. It is aresiduated lattice (in G. Birkhoff’s terminology) because (a : b) (equalto the largest lattice element x such that [b, x] ≤ a) always exists.The concept of a lattice ordered monoid, which arose naturally in idealtheory, has been studied by Krull, Birkhoff, Ward and Dilworth, andnumerous others. (See Birkhoff’s Lattice Theory, Ch. XIV, especiallySection 8 on commutation lattices.)

The theory presented in this book does not lie in the tradition ofthese ongoing axiomatic studies, although it may throw some new lighton the subject. The worth of our theory stems, rather, from its verybroad applicability combined with depth. We shall be working within abroad class of algebras of diverse kinds, with a general definition of com-mutator which turns out to determine a unique operation over everyalgebra. The commutator proves to be an excellent tool for arriving atdeep algebraic results in this very general setting. The first commutatortheory approaching this level of generality was created by the Englishmathematician J. D. H. Smith, and presented in his 1976 book Mal’cevVarieties. Smith dealt with an important subclass of the congruencemodular varieties, namely the varieties in whose algebras all congru-ences permute, which he called Mal’cev Varieties. His theory rapidlyevolved into the theory we shall present. The work was done chiefly bythe German mathematicians J. Hagemann and C. Herrmann, and pre-sented in their papers Hagemann-Herrmann [44] and Herrmann [45].Many important details and simplifications were worked out later byother people, notably by H. P. Gumm and W. Taylor.

General remarks. The operation we shall study is a binary op-eration defined for pairs of congruences 〈θ, ψ〉 of an algebra and givinganother congruence [θ, ψ]. This operation can be easily defined forcongruences of any algebra, but it seems to be especially well behavedonly for algebras in congruence modular varieties. Two of its propertiescontribute most obviously to its strength in applications.

LetA be an algebra in a congruence modular variety, L be its latticeof congruences and θ and ψ be two members of L. If the commutatorof θ and ψ is as large as the theory allows it to be – that is, if it isidentical with the intersection of θ and ψ – then in the neighborhood ofthis pair of congruences L is very like a distributive lattice. Wheneverψ ≤ ψ1 ∨ ψ2, for example, then θ ∧ ψ ≤ (θ ∧ ψ1) ∨ (θ ∧ ψ2) (where∨ and ∧ denote join and meet in L). There is a sophisticated way tostate these facts. Consider this binary relation on L: θ ∼ ψ if and onlyif θ ∨ ψ is solvable over θ ∧ ψ. This is an equivalence relation and itrespects commutators and lattice joins and meets. The quotient lattice

INTRODUCTION 3

L/ ∼ is a distributive commutation lattice in which [α, β] = α∧β. Wemay remark that at one extreme end of the spectrum of congruencemodular varieties, the congruence distributive varieties are preciselythose in which the commutator is identical with the intersection in allcongruence lattices.

The second important property concerns Abelian congruences. Ina quotient algebra B = A/[θ, ψ], the congruence β = θ ∧ ψ/[θ, ψ]is an Abelian congruence, that is [β, β] = 0. This implies that eachequivalence class of β is an Abelian group and every operation of thealgebra B, with its variables restricted to range over fixed equivalenceclasses of β is a group homomorphism (plus a constant). In the extremecase, θ = ψ = 1A (= A×A) and [1A, 1A] = 0A. The algebra A is thenpolynomially equivalent to a module over a ring.

Our presentation of commutator theory is essentially different fromearlier published accounts. Hagemann and Herrmann defined the com-mutator of two congruences on an algebra A, and derived its prop-erties, by applying the modular law in a quite nontrivial fashion tocongruences on subalgebras of A, A2 and A3. A congruence of A isitself an algebra (a subalgebra of A2), and in fact they dealt mainlywith congruences on congruences of A. Gumm’s approach to commu-tator theory (in [38], [40] and [40]) is heavily flavored with geometricanalogies developed in his earlier papers [36] and [39], and is closer inspirit to what Smith did. Gumm made good use of A. Day’s derivedoperations, whose existence for a variety is equivalent to congruencemodularity. Our path to commutator theory is more direct than eitherof these. We define the commutator in any algebra by considering theset of all term operations of the algebra. Then assuming that algebrabelongs to a congruence modular variety, we use Day’s operations topresent a set of generators for [θ, ψ]. Following this result, the basicproperties of commutators are easily derived.

This book is an expanded version of a manuscript entitled Thecommutator, an overview that we used as the text for a week longworkshop on commutator theory held at the Puebla Conference onUniversal Algebra and Lattice Theory in January 1982.

The first chapter of the present manuscript uses groups and ringsto motivate the definition of the commutator. The second chaptergives a brief introduction to the results and definitions of universalalgebra which will be required later. It also gives a proof of Day’scharacterization of varieties with modular congruence lattices. In thecourse of the proof we prove two useful lemmas. The third chapter givesseveral definitions of the commutator and proves some simple resultswhich do not depend on congruence modularity. The fourth chapter

4 INTRODUCTION

shows that all of these definitions of the commutator are equivalent incongruence modular varieties and proves some of the fundamental factsabout the commutator.

The fifth chapter proves Herrmann’s fundamental result on Abelianalgebras and introduces Gumm’s 3-ary term and his characterizationof Abelian congruences. The sixth chapter proves various permutabil-ity theorems and gives Gumm’s new Mal’cev condition for congruencemodularity. The seventh chapter gives several results on nilpotent al-gebras, including a theorem describing the structure of such algebras.A result used in Burris-McKenzie [8], whose proof has not appearedin print before, is included. It is Proposition 7.8 (which was 8.7 inBurris-McKenzie).

Chapter 8 contains some results relating to congruence identities.These are identities in the join, meet and commutator operations sat-isfied by the congruence lattices. This subject is in its infancy and theresults presented are far from representing a complete theory.

The blocks of an Abelian congruence of an algebra A in a (modu-lar) variety V behave very much like modules over a ring built from thepolynomials of A. Chapter 9 defines rings naturally associated with V.In our 1981 manuscript we only considered one ring for each V, whichwe denoted R(V). In the present version we extend this concept in twoways. First we allow constants so that the ring is built from polynomi-als, not just terms. Secondly we form “matrix” rings. In this mannerwe are able to capture the polynomial action between different blocksof an Abelian congruence, not just the action within one block. Somestrong structural connections between the module associated with anAbelian congruence and the algebra are proved. In the case that thevariety consists solely of Abelian algebras the connection is especiallystrong. The last part of the chapter establishes the exact connection.In this case the algebra is polynomially equivalent to the module in avery specific way.

Chapter 10 uses the commutator theory to prove a variety of prob-lems on structure and representation in modular varieties. The firstsection presents some generalizations of Jonsson’s theorem to modu-lar varieties. The second section proves various stronger results forfinitely generated varieties. The third section uses some of the previ-ous material and some results from Chapter 9 to prove the authors’characterization of finite algebras which generate a residually small va-riety. The fourth section defines the concept of chief factors and showsthat if A is a finite algebra then the chief factors which occur in thevariety generated by A have size bounded by |A|. As a corollary we

INTRODUCTION 5

have that the simple algebras in the variety generated by A have sizeat most |A|.

Chapter 11 deals with joins and products of modular varieties. Itis shown that for two such varieties, V0 and V1, if the subdirectlyirreducible algebras in Vi have cardinality at most λi, i = 0, 1, then thesubdirectly irreducible algebras in the join of those varieties have thecardinality at most λ0λ1. A generalization of a theorem of Herrmannon products of varieties is also given.

A finite simple algebra having no nontrivial subalgebras is calledstrictly simple. Chapter 12 deals with varieties generated by such al-gebras. We use the commutator theory to give relatively easy proofsof some of the results of D. Clark and P. Krauss. Chapter 13 studiesMal’cev conditions corresponding to the satisfaction by congruence lat-tices of certain lattice identities. Previously only the distributive law,the modular law, and the two trivial identities were known to have aMal’cev condition. This chapter displays an infinite class of identitiesdefined by Mal’cev conditions.

M. Vaughan-Lee has shown that a finite nilpotent algebra of primepower order having an equationally defined constant has a finite basisfor its identities (assuming, of course that there are only finitely manybasic operations). Chapter 14 proves this without the assumption thatthere is an equationally defined constant.

Exercises are given at the ends of most chapters. A great dealof important material is included in these exercises and the readershould at least glance at them. They include interesting examplesand applications, as well as significant theorems. We have tried toavoid leaving routine calculations as exercises, and indeed some of theexercises require a substantial amount of ingenuity. But do not worry,you can always cheat: all of the solutions appear in the back. Becauseof this we have felt justified in using the results of the exercises.

The authors wish to thank C. Bergman, G. Birkhoff, A. Day, Heinz-Peter Gumm, W. Lampe and J. B. Nation for the interest and assis-tance they rendered us at several stages in the preparation of thismanuscript.

CHAPTER 1

The Commutator in Groups and Rings

Let G be a group and let L(G) denote its lattice of normal sub-groups. Suppose that M, N ∈ L(G). The group commutator (commu-tator of M and N), written as [M,N], is the subgroup of G generatedby all elements [m,n] = m−1n−1mn, with m ∈ M and n ∈ N. Itis a normal subgroup of G. We thus have a binary operation, calledcommutator, acting on normal subgroups. (The usual operations ofa lattice are join and meet, here denoted by ∨ and ∧.) If π : G → G′

is an onto homomorphism, then L(G′) is naturally isomorphic to aninterval in L(G) consisting of the members that lie above the kernelof π. We can easily determine commutators of normal subgroups inG′ if we know the commutators of their inverse images in G. (See (2)below.) Thus we are going to consider the commutator as an operationnot just in one lattice L(G), but rather as a global operation, definedat once in all such lattices. The following properties are easily verifiedfor M, N, Ni ∈ L(G), i ∈ I, and π : G ։ G′ a surjection.

[M,N] ⊆M ∩N(1)

[π(M), π(N)] = π([M,N])(2)

[M,N] = [N,M](3)

[M,∨

i∈I

Ni] =∨

i∈I

[M,Ni](4)

H = [M,N] is the least normal subgroup of Gsuch that in G/H every element ofM/H commutes with every elementof N/H.

(5)

The next property is not quite so obvious.

(6)The commutator is the greatest binary operationdefined on L(G) for every group G and satisfying(1) and (2).

7

8 1. THE COMMUTATOR IN GROUPS AND RINGS

To prove this, we suppose that C is another binary operation de-fined in L(G) for every groupG and that C satisfies (1) and (2). LetM,N ∈ L(G) for some group G. We shall show that C(M,N) ⊆ [M,N].To do this, we define four subgroups of G×G:

G(M) = {〈x, y〉 : x ∈ G, x−1y ∈M}

∆ = {〈x, y〉 : x ∈ N, x−1y ∈ [M,N ]}

B = {〈x, 1〉 : x ∈ [M,N ]}

M1 = {〈x, 1〉 : x ∈M}.

Now one can verify that ∆, B, M1 ∈ L(G(M)). [This we leave to thereader. The fact that ∆ ∈ L(G(M)) is what makes the proof work.]The projection homomorphism π of G(M) at the first coordinate mapsG(M) homomorphically onto G. Moreover, we obviously have that

π(∆) = N

π(B) = [M,N]

π(M1) = M

From the fact that C satisfies (1) we get that

C(M1,∆) ⊆M1 ∧∆ ⊆ B.

To finish the argument we use (2):

C(M,N) = π(C(M1,∆)) ⊆ π(B) = [M,N].

In the most general varieties of algebras, there is no such thing asa normal subgroup. The kernels of homomorphisms must be identifiedwith congruences. For such a variety, we can replace normal subgroupsby congruences in the formulation of (1) and (2), and consider binarycongruence operations defined globally (in the congruence lattice ofevery algebra). It is quite easy to see that there must be a greatestsuch operation satisfying (1) and (2). We shall not be able to sayvery much in general about that operation. However, for congruencemodular varieties we shall be able to say a great deal. But we need abetter way to come to grips with the operation. Our approach in thisbook is to find a formulation of the property (5) which is capable ofgeneralization to all algebras, and use that to build a more constructivedefinition of a general commutator. Fortunately, this idea works outbeautifully for congruence modular varieties. We get a commutatordefined by an abstraction of (5), which satisfies (1) and (2), in fact is

1. THE COMMUTATOR IN GROUPS AND RINGS 9

the greatest operation satisfying (1) and (2), and which also satisfies (3)and (4).

Our constructive definition of a commutator occurs in the thirdchapter. It is rather hard to see statement (5) behind that definition.The most we can do to motivate it is to work through the equivalenceof the two for groups. That is what we shall now do.

Let G be a group and let M, N ∈ L(G) be such that [M,N] = {1}.This means of course that every member of M commutes with everymember of N. It implies that the term operations of the group, whenapplied to elements ofM∪N , take a simple form. Let t(x1, . . . , xp, y1, . . . , yq)be any term operation of p + q variables. It can be expressed in theform

t(x1, . . . , xp, y1, . . . , yq) = u1 . . . uk

where each ui ∈ {x1, x−11 , . . . , xp, x

−1p , y1, y

−11 , . . . , yq, y

−1q }. Now if m =

〈m1, . . . , mp〉 ∈ Mp and n = 〈n1, . . . , nq〉 ∈ Nq, then since mi com-mutes with nj , we have

t(m,n) = r(m) · s(n)

for a certain pair of term operations r(x) and s(y) which depend onlyon the term t (and not on any other parameters of the situation we areconsidering).

Now let t(x,y), r(x) and s(y) be terms related as above, and letm1, m2 ∈ Mp and n1, n2 ∈ Nq. By substituting these elements forthe variables in t(x,y), we obtain four elements of the group, and wearrange them in the form of a matrix

(5′)

[

t(m1,n1) t(m1,n2)t(m2,n1) t(m2,n2)

]

=

[

t11 t12t21 t22

]

Note that the elements in each column of this matrix belong to onecoset of M, and the elements in each row of this matrix belong to onecoset of N. Since

[

t11 t12t21 t22

]

=

[

r(m1) · s(n1) r(m1) · s(n2)r(m2) · s(n1) r(m2) · s(n2)

]

we have the following bi-implications:

t11 = t12 ↔ t21 = t22

t11 = t21 ↔ t12 = t22(5′′)

It is this simple property that we shall exploit to make the generaldefinition of the commutator. Our assumption that [M,N] = {1}

10 1. THE COMMUTATOR IN GROUPS AND RINGS

implies that (5′′) holds for every term operation t(x,y). Conversely, bychoosing t(x, y) = x−1yx, we get matrices of the form

[

t(m,n) t(m, 1)t(1, n) t(1, 1)

]

=

[

m−1nm 1n 1

]

and so (5′′) for this term implies that [M,N] = {1}. The condition(5′′) has been called the term condition.

Now we turn to rings and consider what the commutator will turnout to be for congruences on rings. Let J and K be ideals of a ring R.If the term condition is satisfied, then by considering matrices derivedfrom the term operations s(x, y) = x · y and t(x, y) = y · x, in fact thematrices[

xy 00 0

]

=

[

xy x00y 00

]

,

[

yx 00 0

]

=

[

yx 0xy0 00

]

x ∈ J, y ∈ K.

we find that J ·K = K · J = {0}. This suggests that we might define[J,K], for any two ideals of a ring, as the ideal generated by J·K∪K·J.

If we do make this definition, then it is easy to modify the argumentwe gave for groups, and prove that [J,K] is the largest binary globaloperation on ideals in the variety of rings which satisfies (1) and (2). Itis also easy to prove that [J,K] = {0} if and only if the term conditionis satisfied.

Exercise

1. Let V be the variety of modules over some ring. Identifycongruences with submodules. Prove that the term condi-tion holds for any two submodules of a module. Prove that[M,N] = {0} is the only binary global operation on submod-ules in V which satisfies (1) and (2).

CHAPTER 2

Universal Algebra

We have to assume that our readers have had a basic introduction touniversal algebra, and are familiar with the most elementary conceptsof modern logic. Excellent references are the books of Birkhoff [5],Crawley and Dilworth [15], and Gratzer [33] and [34], (for everythingpertaining to algebras and lattices) and the book of Burris and Sankap-panavar [10] (for the logic and everything to do with varieties). Withthe aid of these references it is possible for a reader unfamiliar withuniversal algebra to read this book. Our purpose in this chapter is toestablish our point of view, and the notation to be used.

The set {0, 1, 2, . . . } of natural numbers is denoted by ω, and itsmembers satisfy n = {0, 1, . . . , n− 1}. We use An to denote the directproduct of n copies of A.

By an algebra we mean any model for a first order language Lwhose nonlogical symbols are operation symbols Fi with arity ρ(i),i ∈ I, where ρ(i) ∈ ω. Thus an algebra consists of a nonvoid set Atogether with a collection {Fi : i ∈ I} of operations on A, i.e., mapsFi from Aρ(i) into A, one for each operation symbol in the language.We use boldfaced capital letters to denote algebras and the same plaincapital letter for the corresponding underlying sets. The operation onthe set A assigned to the operation Fi in an algebra A will be denotedFAi ; and thus an algebra A has the form 〈A, FA

i (i ∈ I)〉. Whenever wedeal with a class of algebras, it will be assumed that all the algebras aremodels for one language, i.e., that they are of the same similarity type.Thus each class K of algebras is attached to a language L(K) = L, andK ⊆ Mod(L), where of course Mod(L) denotes the class of all models ofL. When A ∈ Mod(L) as above, then FA

i is called the interpretationof Fi in A, and is a ρ(i)-ary operation on the set A. The type of A isthe function ρ and A has finite type if and only if the domain of ρ,that is I, is finite. Constants are the same as 0-ary operations. By theuniverse of an algebra A we mean the underlying set A of the algebra.The cardinality of A is denoted by |A|.

The set of terms in the variables vi, i ∈ ω, for a language L is thesmallest set T containing the vi and such that if t1, . . . , tn ∈ T and

11

12 2. UNIVERSAL ALGEBRA

Fi is an operation symbol in L with ρ(i) = n, then Fi(t1, . . . , tn) ∈ T .If A ∈ Mod(L) is an algebra then to each term of L we can asso-ciate an operation on A in an obvious way. Thus if t(v0, . . . , vn−1)is a term of L in which only the (distinct) variables v0, . . . , vn−1 ap-pear, then tA denotes the corresponding n-ary operation on A. Suchoperations are called term operations, or in the older literature, de-rived operations. The set of term operations of A can also be de-fined as the smallest set of operations that is closed under composi-tion and contains the basic operations {FA

i : i ∈ I} and the trivialprojection operations πni (where πni (x0, . . . , xn−1) = xi). The polyno-mial operations of A constitute the smallest set that is closed undercomposition and contains the basic operations of A, the projectionoperations, and the constant 0-ary operations on A. For each poly-nomial operation f(x0, . . . , xn−1) of A, there exists a term operationp(x0, . . . , xm−1) for some m ≥ n and elements an, . . . , am−1 of A, suchthat f(x0, . . . , xn−1) = p(x0, . . . , xn−1, an, . . . , am−1) holds identically.The expression algebraic function is often used in the literature to referto what we call a polynomial operation.

AlgebrasA and B are called equivalent if and only if they have thesame universe and exactly the same set of term operations. (Equivalentalgebras need not be of the same type.) Algebras A and B are calledpolynomially equivalent if and only if they have the same universeand exactly the same polynomial operations.

Suppose that A ∈ Mod(L) and that s = s(v0, . . . , vn−1) and t =t(v0, . . . , vn−1) are terms of L. The formula s ≈ t is called an equation.We write A |= s ≈ t to denote that sA = tA, i.e., sA(a0, . . . , an−1) =tA(a0, . . . , an−1) for all a0, . . . , an−1 ∈ A. When A |= s ≈ t we saythat s ≈ t is an identity of A. We sometimes use the terminologylaw or equation to mean the same as identity. K |= s ≈ t means thatA |= s ≈ t for all A ∈ K.

If Σ is a set of equations of L then

Mod(Σ) = {A ∈ Mod(L) : A |= ε for all ε ∈ Σ}

Classes of the form Mod(Σ), Σ a set of equations of L are called vari-eties (or equational classes). By a theorem of G. Birkhoff [4], a classV ⊆ Mod(L) is a variety if and only if V is closed under the formationof homomorphic images, subalgebras and products. The smallest vari-ety containing a class K ⊆ Mod(L) is identical with HSP(K), whereH , S and P are the operators which close classes under homomorphicimages, subalgebras and direct products, respectively. We interpretthese operators in such a way that the closure of K under any of themcontains all isomorphic copies of its members. V ⊆ Mod(L) is a variety

2. UNIVERSAL ALGEBRA 13

if and only if V = HSP(V). We let V denote the operator HSP, i.e.,V (K) = HSP(K).

A congruence of an algebra A is an equivalence relation on theuniverse of A that is induced by some homomorphism with domain A,i.e., two elements of A are related by the congruence if and only if theyhave the same image under the homomorphism. The congruences ofan algebra A constitute a lattice (ordered by set inclusion) which wedenote by Con A. The lattice operations of Con A will be denotedby ∨ for join, and ∧ for meet, with

∨

and∧

for the infinitary versions.The least and greatest elements of the lattice will be denoted by 0 and1 (or 0A and 1A for sake of clarity). If θ is in ConA, then the set ofequivalence classes of θ can be made, in a natural way, into an algebrawhich we denote A/θ. For x ∈ A we let x/θ denote the equivalenceclass containing x. The map x → x/θ is a natural homomorphismfrom A onto A/θ. The notations x ≡ y (mod θ) and x θ y both signifythat 〈x, y〉 ∈ θ. A similar notation with an entirely different meaningis α/β = {γ ∈ ConA : β ≤ γ ≤ α}. α/β is called a quotient orinterval in the lattice Con A.

Congruences on A can alternatively be characterized as the equiv-alence relations R on A such that R is a subalgebra of A ×A. Con-gruences, in general, do not behave so well as they do in groups, ringsand modules; for instance two congruences on A may share a block,x/α = x/β, for some x, without being identical.

We call an algebra simple if it has exactly two congruences, andsubdirectly irreducible if its congruence lattice has exactly oneatom, called the monolith and usually denoted by µ, such that µ ≤ θfor every nonzero congruence θ. Another theorem of Birkhoff tells usthat every algebraA is isomorphic to a subdirect product of subdirectlyirreducible homomorphic images Ai, i ∈ I. That is, A is isomorphicto an algebra C ≤

∏

i∈I

Ai such that C projects onto each factor Ai.

An element c of a complete lattice is compact if c ≤∨

X for somesubset X of the lattice implies that c ≤

∨

F for some finite F ⊆ X .A complete lattice is compactly generated or algebraic if every el-ement is the join of compact elements. Congruence lattices are alwayscompactly generated. An element q of a lattice is completely meetirreducible if q =

∧

Y implies q ∈ Y . An algebra A is subdirectlyirreducible if and only if the least element of Con A is completelymeet irreducible. Since if θ ∈ ConA, Con(A/θ) is isomorphic to theinterval 1/θ in Con A, representations of A as a subdirect product ofsubdirectly irreducible algebras correspond to sets of completely meet


irreducible congruences of A which meet to 0. For compactly gener-ated lattices, it is not hard to prove that every element is the meet ofcompletely meet irreducible elements.

A lattice is distributive if it satisfies the distributive law, x∧ (y ∨z) ≈ (x ∧ y) ∨ (x ∧ z). It is modular if it satisfies the implicationx ≥ y → x ∧ (y ∨ z) ≈ y ∨ (x ∧ z). In a modular lattice the notion oftransposed quotients is very important. We say that α/β transposesdown onto γ/δ (where α ≥ β and γ ≥ δ, of course), and we writeα/β ց γ/δ if β ∧ γ = δ and β ∨ γ = α. We also write γ/δ ր α/β forthis concept.

The congruence on A generated by X , a subset of A × A, is de-noted CgA(X), or sometimes Cg(X), and we abbreviate Cg({〈x, y〉})as Cg(x, y). The subalgebra of A generated by Y is denoted SgA(Y ) orSg(Y ). Boldface symbols like x denote n-tuples of elements. Hom(A,B)denotes the set of homomorphisms from A to B; this can be the emptyset. If f ∈ Hom(A,B) then the kernel of f (i.e., the induced congru-ence on A) is written kerf . If B is a subalgebra of A and f is a mapwith domain A, f |B denotes the restriction of f to B. If θ ∈ ConA,θ|B denotes the restriction of θ to B, i.e., θ|B = θ ∩ (B × B).

A variety V is called distributive if for all A ∈ V, Con A isdistributive; modular if for all A ∈ V, Con A is modular; permutableif for all A ∈ V and for all θ, φ ∈ ConA, θ ◦φ = φ◦θ. (Here θ ◦φ is therelational product: {〈x, y〉 : ∃z ∈ A, x θ z φ y}. The equality impliesθ ◦ φ = θ ∨ φ, and the condition implies that ConA is modular.)

Let V be a variety which contains an algebra with at least twoelements and X be a nonvoid set. There is an algebra F ∈ V (uniqueup to isomorphism) having these properties:

(1) F contains X and is generated by X ;

(2) whenever A ∈ V and f maps X into A, then there exists a

unique homomorphism f : F → A such that f ⊇ f (i.e.,

f |X = f).

Such an algebra F is said to be a free algebra in V, freely generatedby X , and is denoted by FV(X). Some other notations may be usedfor denoting free algebras, such as FV(λ) (if |X| = λ), or FV(x, y, z) (ifX = {x, y, z}). An important fact about free algebras is that for an

algebra A, FV (A)(λ) is a subalgebra of A|A|λ (see Birkhoff [5]). Thusif A and λ are both finite, then so is FV (A)(λ). Consequently, if A is afinite algebra then V (A) is locally finite, i.e., every finitely generatedalgebra in V (A) is finite.


The next theorems characterize permutability, distributivity, andmodularity of the congruence lattices of the algebras in a variety bythe existence of terms satisfying certain equations. Conditions of thisform are known as Mal’cev conditions.

Theorem 2.1. For any variety we have:

(1) (Mal’cev [61]) V is permutable if and only if there is a termp(x, y, z) in the language of V such that the following equationsare valid in V.

p(x, x, y) ≈ y ≈ p(y, x, x)

(2) (Jonsson [54]) V is distributive if and only if for some n thereare terms d0(x, y, z), . . . , dn(x, y, z) such that V satisfies

(i) d0(x, y, z) ≈ x, dn(x, y, z) ≈ z,

(ii) di(x, y, x) ≈ x, i ≤ n,

(iii) di(x, x, y) ≈ di+1(x, x, y), for all even i < n,

(iv) di(x, y, y) ≈ di+1(x, y, y), for all odd i < n.

Proof. The proof of (2) will be omitted since it is similar to theproof of Theorem 2.2 below and we will not use (2). To sketch (1) firstsuppose V has such a term p and that x θ y ψ z, for some θ, ψ∈ ConA,where A ∈ V. Then x = p(x, y, y) ψ p(x, y, z) θ p(x, x, z) = z, showingthat θ and ψ permute. For the converse suppose that V is permutableand let FV(x, y, z) be the free V–algebra generated by x, y and z. Letθ = Cg(x, y) and ψ = Cg(y, z). Then x θ y ψ z, so there is anelement p in FV(x, y, z) with x ψ p θ z. Let p(x, y, z) be a term whichrepresents p, that is, pF(x, y, z) = p in FV(x, y, z) where pF(x, y, z)is the interpretation of p as a term operation of FV(x, y, z). Let σbe the endomorphism of FV(x, y, z) which maps x to x, y to y andz to y. One easily checks that ψ is the congruence associated withthis endomorphism. Since σ is the identity map on the subalgebragenerated by x and y, x ψ pF(x, y, y) implies x = pF(x, y, y). Similarlyz = pF(x, x, z). Of course these facts holding in the free algebra implythat the equations of (1) hold in V. �


Theorem 2.2. (Day [19]) A variety V is modular if and only iffor some n there are terms m0(x, y, z, u), . . . , mn(x, y, z, u) such that Vsatisfies

(i) m0(x, y, z, u) ≈ x, mn(x, y, z, u) ≈ u

(ii) mi(x, y, y, x) ≈ x, i ≤ n

(iii) mi(x, x, y, y) ≈ mi+1(x, x, y, y), for all even i < n

(iv) mi(x, y, y, z) ≈ mi+1(x, y, y, z), for all odd i < n

Lemma 2.3. Let V be a variety having terms mi(x, y, z, u), i =0, . . . , n, satisfying the conditions of Theorem 2.2, and let A ∈ V,γ ∈ ConA, a, b, c, d ∈ A with 〈b, d〉 ∈ γ. Then 〈a, c〉 ∈ γ if and onlyif for all i ≤ n, mi(a, a, c, c) γ mi(a, b, d, c).

Lemma 2.4 (The Shifting Lemma). Let V be a modular varietyand let A ∈ V and ψ, θ0, θ1 ∈ ConA. Suppose that a, b, c, d ∈ A,〈a, b〉, 〈c, d〉 ∈ θ0, 〈a, c〉, 〈b, d〉 ∈ θ1, and ψ ≥ θ0 ∧ θ1. Then 〈b, d〉 ∈ ψimplies 〈a, c〉 ∈ ψ. Pictorially,

implies

θ0

θ1 ψ ψ

a

c

b

d

a

c

b

d

Note. In diagrams like this, lines without labels are implicitly as-sumed to be labeled by (the endpoints are congruent modulo) any labelappearing on a parallel line. Terms satisfying the requirements of The-orem 2.2 are called Day terms.

Proofs. Consider the following conditions for a variety V.

(1) V is modular.

(2) V has Day terms.

(3) V has termsmi(x, y, z, u), i = 0, . . . , n, satisfyingmi(x, y, y, x) ≈x, such that if A ∈ V, γ ∈ ConA, a, b, c, d ∈ A, with 〈b, d〉 ∈γ, then 〈a, c〉 ∈ γ if and only if for all i ≤ n, mi(a, a, c, c) γmi(a, b, d, c).


(4) V satisfies the conclusion of the Shifting Lemma, i.e., if a, b,c, d ∈ A ∈ V and ψ, θ0, θ1 ∈ ConA with ψ ≥ θ0 ∧ θ1, thenthe implication of the figure holds.

We let (4′) be the condition identical to (4) except that we assumeonly that θ0 is a reflexive, compatible relation (rather than θ0 a con-gruence). Clearly (4′) → (4). To prove the theorem and the lemmasit suffices to show that (1) → (2) → (3) → (4′) → (1). This will alsoshow that condition (3) is equivalent to modularity. After the proof wewill outline a proof that (4) is also equivalent to modularity.

(1) → (2). Assume V is modular. Let F = FV(x, y, z, u) be thefree V–algebra on four generators, and let α = Cg(x, u) ∨ Cg(y, z),β = Cg(x, y)∨Cg(z, u), and γ = Cg(y, z) be in Con F. Then (x, u) ∈α ∧ (β ∨ γ). Since V is modular, 〈x, u〉 ∈ γ ∨ (α ∧ β). This means thatfor some n there are elements w0 = x, w1, . . . , wn = u of FV(x, y, z, u)such that wi α ∧ β wi+1 if i is even, and wi γ wi+1 if i is odd. Letx = m0(x, y, z, u), m1(x, y, z, u), . . . , mn(x, y, z, u) = u be the termsrepresenting w0, w1, . . . , wn, i.e., wi = mF

i (x, y, z, u). Since γ ≤ αall of the wi’s are in the same α class. Thus x α mF

i (x, y, z, u) αmFi (x, y, y, x). Since α restricted to the subalgebra generated by x and

y is trivial, x = mFi (x, y, y, x). Hence x ≈ mi(x, y, y, x) holds in V.

Similar arguments show that the other equations of Theorem 2.2 hold.(2) → (3). Let V be a variety having Day terms and let A ∈ V,

γ ∈ ConA, a, b, c, d ∈ A, with b γ d. First assume a γ c. Then in Ami(a, a, c, c) γ mi(a, a, a, a) = a, and mi(a, b, d, c) γ mi(a, b, b, a) = a.For the converse, setting ui = mi(a, a, c, c) and vi = mi(a, b, d, c), andassuming ui γ vi for all i ≤ n, we can show ui γ ui+1 for all i < n; thusa = u0 γ un = c. Namely, for even i < n, ui = ui+1; and for odd i < n,ui γ vi γ mi(a, b, b, c) = mi+1(a, b, b, c) γ vi+1 γ ui+1.

(3) → (4′). Assume that V is a variety satisfying (3) and thatthe conditions implied by the left hand picture of Lemma 2.4 hold inalgebra A ∈ V, where θ1 and ψ are congruences on A and θ0 is areflexive, compatible relation. Then for any i ≤ n, mi(a, a, c, c) θ0mi(a, b, d, c) and mi(a, a, c, c) θ1 mi(a, a, a, a) = a = mi(a, b, b, a) θ1mi(a, b, d, c). Hence, mi(a, a, c, c) θ0 ∧ θ1 mi(a, b, d, c). Now the conclu-sion of Lemma 2.4 follows by applying (3) with γ = ψ, since ψ ≥ θ0∧θ1and (b, d) ∈ ψ.

(4′)→ (1). Suppose that V satisfies (4′) and that A ∈ V and α, β,γ ∈ ConA with α ≥ γ. We need to show that α∧ (β∨γ) = (α∧β)∨γ.The left side may be written as

⋃

n<ω

(α∩Rn), where R0 = β and Rk+1 =

Rk ◦ γ ◦ β. Thus it suffices to show that α ∩Rn ⊆ (α ∧ β) ∨ γ. This isclear for n = 0. Assume it is true for n = k. Let 〈a, b〉 ∈ α ∩ Rk+1 =


α∩(Rk◦γ◦β). Then there are elements c and d inA with a Rk c γ d β b.Thus since β ⊆ Rk we have

Rk

α γ ∨ (α ∧ β)a

b

c

d

Now we conclude from (4′) that 〈a, b〉 ∈ γ ∨ (α∧ β), as desired. �

Lemma 2.4 was discovered by H. P. Gumm, who also named it. It isquite important in the development of the theory of modular varietiesand plays a central role in his geometric approach. For a variety, thevalidity of the Shifting Lemma is equivalent to modularity, as Gummshowed [41]. The above proof shows the Shifting Lemma holds inmodular varieties. Or alternatively we can argue directly that if θ0 ∧θ1 ≤ ψ and the conditions implied by the left hand figure of the ShiftingLemma hold then (a, c) ∈ θ1∧ (θ0∨ (θ1∧ψ)) = (θ0∧ θ1)∨ (θ1∧ψ) ≤ ψ.Conversely, if condition (4) holds in a variety V, let FV(x, y, z, u) be thefree V–algebra and let α = Cg(x, u)∨Cg(y, z), β = Cg(x, y)∨Cg(z, u),γ = Cg(y, z). Then

β

α γ

x

u

y

z

Thus 〈x, u〉 ∈ γ ∨ (α ∧ β). Just as in the proof that (1)→ (2), thissituation implies the existence of terms mi which satisfy the equationsin Theorem 2.2, which in turn imply modularity.

As we mentioned above, terms satisfying the equations of Theo-rem 2.2 will be called Day terms. A term satisfying the condition ofTheorem 2.1(1) will be called a Mal’cev term, and of course termssatisfying Theorem 2.1(2) will be called Jonsson terms.

To underline the power of Lemma 2.3 we shall now derive one ofGumm’s interesting geometrical results. Another one of his geometricalresults is contained in the exercises. Neither of these results will be usedin the sequel. We assume here that A is an algebra such that V (A) ismodular.

Theorem 2.5. (The Cube Lemma) Suppose that θ0, θ1, ψ ∈ConA and ψ ≥ θ0 ∧ θ1. Let a, b, c, d, a′, b′, c′, d′ be elements of A.Then

EXERCISES 19

impliesa

b

c

d

a′

b′

c′

d′θ1

θ0

ψ

a

b

c

d

a′

b′

c′

d′θ1

θ0

ψ

ψ

Proof. One can readily verify the congruences of the followingdiagram, where i ≤ n.

θ0

θ1 ψ

mi(a, b, d, c) mi(a′, b′, d′, c′)

mi(a, a, c, c) mi(a′, a′, c′, c′)

By the Shifting Lemma, mi(a, b, d, c) ψ mi(a, a, c, c). Now the resultfollows from Lemma 2.3. �

Exercises

1. A quasigroup is an algebra 〈A, ·, /, \〉 with three binary oper-ations which satisfy the laws

x · (x\y) ≈ y (x/y) · y ≈ x

x\(x · y) ≈ y (x · y)/y ≈ x

A loop is a quasigroup with a constant 1 satisfying the law1 · x ≈ x · 1 ≈ x. Show that left multiplication by an elementof a quasigroup A defines a bijection from A to A. Show thatif A is a finite quasigroup, then there is a binary term t(x, y)which uses only multiplication, such that x/y = t(x, y).

2. Show that if 〈A, ·, /, \〉 is a quasigroup then

p(x, y, z) = (x/(y\y)) · (y\z)

andq(x, y, z) = ((x · y)/x)\(x · z)

are both Mal’cev terms for A. Hence the class of all quasi-groups forms a permutable variety.

3. (A. Day) Let V be a permutable variety with Mal’cev termp(x, y, z). Show that V is modular with Day termsm0(x, y, z, u) =x, m1(x, y, z, u) = p(x, p(x, y, z), u), m2(x, y, z, u) = u.


4. Prove that the following form of the Shifting Lemma, pre-sented geometrically, is equivalent to that given in Lemma 2.4,where ψ′ = ψ ∨ (θ0 ∧ θ1).

implies

θ0

θ1 ψ ψ

a

c

b

d

a

c

b

d

ψ′

θ0

θ1

5. Use Lemma 2.3 and Lemma 2.4 to prove the following geomet-rical result of Gumm which is called the Little DesarguesianTheorem. Let θ, ψ, α1, α2 ∈ ConA with θ ∧ αi ≤ ψ, i = 1, 2,and let a, b, c, d, e, f ∈ A. Then

implies

θ

ψ

α1α2

e

ab

c

f

d

ab

e

c

f

d

θα1

α2ψ ψ

CHAPTER 3

Several Commutators

In this chapter V denotes any variety, not assumed to be modular.The first “commutator” is defined with reference to V, so its value maydepend not only on the algebra A but also on the particular variety towhich A belongs, under consideration. The next two “commutators”depend just on A. (All three will turn out to be identical for modularvarieties.)

Definition 3.1. C(V) is the largest binary operation defined onCon A for every A ∈ V and satisfying (for any A,B ∈ V, and θ, ψ ∈ConA, and f ∈ Hom (A,B) a surjective homomorphism with kernel π)

C(V)(θ, ψ) ≤ θ ∧ ψ and

C(V)(θ, ψ) ∨ π = f−1C(V)(f(θ ∨ π), f(ψ ∨ π)).(1)

A rigorous justification of this definition would involve us in con-siderations of axiomatic set theory rather far removed from the subjectat hand. C(V) is the “pointwise join” of all binary congruence opera-tions defined globally over V and satisfying the conditions. (Note thatC(θ, ψ) = 0 is such an operation, so there is at least one.) The otherdefinitions of commutator are completely constructive.

The next definitions are motivated by fact (5′′) of Chapter 1 (aproperty of the commutator for groups).

Definition 3.2. Let α, β, and δ be in ConA.

(1) M(α, β) is the set of all matrices[

t(a1,b1) t(a1,b2)t(a2,b1) t(a2,b2)

]

where ai, i = 1, 2 is a sequence of n elements of A, bi is asequence of m elements of A, m,n ≥ 0, satisfying a1

k α a2k and

b1j β b2

j for k < n and j < m, and t is a n +m variable termoperation on A.

(2) We say α centralizes β modulo δ, and write C(α, β; δ), provided

that for every

[

u11 u12u21 u22

]

in M(α, β), u11 δ u12 implies u21 δ

u22.

21

22 3. SEVERAL COMMUTATORS

(3) [α, β] is the smallest δ for wich C(α, β; δ) holds.(4) [α, β]s is the smallest δ for which both C(α, β; δ) and C(β, α; δ)

hold.1

Obviously C(α, β;α ∧ β) holds and if C(α, β; δi) holds for i ∈ I,then C(α, β;

∧

i∈I δi) holds. So the definitions make sense.We will occasionally use the terminology δ satisfies the α, β–term

condition when α centralizes β modulo δ. Another terminology forthis situation is α annihilates β modulo δ. The origin of this termi-nology is ring theory: If I and J are ideals of a ring, [I,J] = 0 meansIJ = JI = 0.

Each member ofM(α, β) presents two demands, a row demand anda column demand. [α, β] is the least congruence satisfying all of therow demands, [β, α] is the least congruence satisfying all of the columndemands, [α, β]s is the least congruence satisfying all demands. Thefollowing facts are obvious.

Proposition 3.3. Let α, β ∈ Con A

(1) M(α, β) is the subalgebra of A4 generated by all matrices ofthe forms

[

a aa′ a′

]

and

[

b b′

b b′

]

where 〈a, a′〉 ∈ α and 〈b, b′〉 ∈ β.

(2)

[

a bc d

]

∈M(α, β) if and only if

[

a cb d

]

∈M(β, α). �

In this completely general setting we can prove almost nothingabout the three congruence operations just defined, except for the mostobvious facts stated below. We expect that under rather weak condi-tions. C(V)(α, β) ≤ [α, β] must hold.

Proposition 3.4. (1) C(V) is symmetric and satisfies 3.1(1).(2) [α, β] ≤ [α, β]s = [β, α]s ≤ α ∧ β.(3) [α, β] and [α, β]s are monotone in both α and β. �

Exercises

1. Show that if α∧β ≤ δ ≤ α or if (α∨δ)∧β ≤ δ then C(α, β; δ)holds.

1This notation differs from the first edition where we used C(α, β) for what ishere denoted [α, β] and [α, β] for what is here denoted [α, β]s. In a modular variety[α, β] = [α, β]s.

EXERCISES 23

2. If C(αi, β; δ) holds for all i, then C(∨

αi, β; δ) holds.

CHAPTER 4

One Commutator in Modular Varieties;

Its Basic Properties

In this chapter V denotes a fixed modular variety with Day termsm0(x, y, z, u), . . . , mn(x, y, z, u) (see Theorem 2.2). We show that thecommutators defined in Chapter 3 are identical for V. The basic prop-erties of the commutator are derived and its behavior with respect tohomomorphisms, subalgebras, and direct products is described. Finallywe introduce the Hagemann-Herrmann definition of the commutatorand show it is equivalent to our definition. We also give the latticetheoretic characterization of the commutator due to these authors.

Recall the definitions ofM(α, β), [α, β]. and [α, β]s from Chapter 3.

Definition 4.1. For α, β ∈ ConA, let X(α, β) be the set of or-

dered pairs 〈mi(a, b, d, c), mi(a, a, c, c)〉 where

[

a bc d

]

∈ M(α, β) and

i ≤ n.

Proposition 4.2. Let α, β, δ ∈ ConA, A ∈ V, where V is modu-lar.

(1) The following are equivalent(i) X(α, β) ⊆ δ(ii) X(β, α) ⊆ δ(iii) C(α, β; δ) holds(iv) C(β, α; δ) holds(v) [α, β] ≤ δ

(2) [α, β] = [α, β]s = Cg(X(α, β)).

Proof. To prove (1) we show first that (iii) ⇒ (i) ⇒ (iv). Thenby exchanging α and β we have (iv) ⇒ (ii) ⇒ (iii), and it follows thatall are equivalent. Assume first that C(α, β; δ) holds. We have to showthat for any term t, for any a1 ≡ a2 (mod α) and b1 ≡ b2 (mod β),

and for the resulting matrix

[

a bc d

]

∈ M(α, β), u ≡ v (mod δ) where

u = mi(a, a, c, c) and v = mi(a, b, d, c). We will show that

[

uv

]

is the

25

26 4. ONE COMMUTATOR; BASIC PROPERTIES

right hand column of a matrix

[

w uw v

]

∈ M(β, α). (Hence 〈u, v〉 ∈ δ as

〈w,w〉 ∈ δ and C(α, β; δ) holds.) In fact

u = mi(t(a1,b1), t(a1,b1), t(a2,b1), t(a2,b1))

v = mi(t(a1,b1), t(a1,b2), t(a2,b2), t(a2,b1))

and if we replace a1 by a2 at its second occurrence in each of theseterms, and a2 by a1 at its second occurrence, then the results are equal(= w = t(a1,b1) by Proposition 2.2(ii)). The term s that gives rise to[

w uw v

]

is

s(x1,x2,y1,y2,y3,y4,y5,y6)

= mi(t(y1,y2), t(x1,y3), t(y4,y5), t(x2,y6)).

To see (i) ⇒ (iv), suppose

[

a bc d

]

∈ M(α, β) and 〈b, d〉 ∈ δ.

Lemma 2.3 implies 〈a, c〉 ∈ δ showing C(β, α; δ) holds. This concludesthe proof of (1), (2) is immediate from (1). �

Proposition 4.3. For congruences θ, ψ, γi (i ∈ I) on A ∈ V wehave

[θ, ψ] = [ψ, θ] ≤ θ ∧ ψ

and[θ,

∨

i∈I

γi] =∨

i∈I

[θ, γi].

Proof. The first statement is by Proposition 3.4(2). To prove thesecond equality, note first that the right side is contained in the leftby monotonicity (i.e. by Proposition 3.4(3)). Denote the right side byα. By Proposition 4.2(1), to get equality we just have to show thatγ centralizes θ modulo α, where γ =

∨

i∈I γi. By Proposition 4.2, γicentralizes θ modulo α for each i ∈ I.

So let

[

u vr s

]

belong to M(θ, γ) and 〈u, r〉 ∈ α. It is clear that

there exist finite sequences x0, . . . , xk, z0, . . . , zk such that

[

xj xj+1

zj zj+1

]

∈

M(θ, γij) for j < k and

[

x0 xkz0 zk

]

=

[

u vr s

]

. The matrices are obtained

from the same term operation that gives

[

u vr s

]

. Thus inductively

〈xj+1, zj+1〉 ∈ α as γij centralizes θ modulo α, and finally 〈v, s〉 ∈ α.This concludes the proof. �

4. ONE COMMUTATOR; BASIC PROPERTIES 27

Proposition 4.4.

(1) If f ∈ Hom (A,B) is surjective, with kernel π, and θ, ψ ∈ConA, then

[θ, ψ] ∨ π = f−1[f(θ ∨ π), f(ψ ∨ π)]

.(2) If B is a subalgebra of A and θ, ψ ∈ ConA, then the restric-

tions to B satisfy

[θ|B, ψ|B] ≤ [θ, ψ]|B.

Proof. For (1), by Proposition 4.3 [θ, ψ] ∨ π = [θ ∨ π, ψ ∨ π] ∨ π.Thus we may assume without loss of generality that θ, ψ ≥ π. Then fcarries a set of generators of [θ, ψ] ∨ π (namely X(θ, ψ)∪ π) onto a setof generators for [f(θ), f(ψ)]. Hence f([θ, ψ]∨ π) = [f(θ), f(ψ)], whichis equivalent to the desired conclusion.

For (2), θ|B centralizes ψ|B modulo [θ, ψ]|B, or argue by generators.�

Proposition 4.5. Let A = Πi∈IAi and θi ∈ ConAi, i ∈ I. Thenthe map

(3) (θi)i∈I → {〈a,b〉 ∈ A2; 〈ai, bi〉 ∈ θi, for all i ∈ I, and ai = bi

for all but finitely many i ∈ I}

is a lattice isomorphism from Πi∈I Con Ai into Con A. Furthermore,this isomorphism preserves the commutator operation. In particular ifA = A0 ×A1, and θi, ψi ∈ ConAi, i = 0, 1, then

[θ0 × θ1, ψ0 × ψ1] = [θ0, ψ0]× [θ1, ψ1].

Moreover if Πi∈Iθi = {〈a,b〉 ∈ A2 : 〈ai, bi〉 ∈ θi, i ∈ I}, then

(4) [Πi∈Iθi,Πi∈Iψi] ≤ Πi∈I [θi, ψi].

Proof. We give an argument avoiding the term condition, in orderto show that conditions Definition 3.1(1) and additivity imply (3). Letg : ΠI Con Ai → Con A be the map defined in (3). We leave it tothe reader to show that g defines a lattice isomorphism of ΠI Con Ai

into Con A. Let λ = {〈a,b〉 ∈ A2 : ai = bi for all but finitely manyi ∈ I}. Let pi be the projection homomorphism A ։ Ai and ηi itskernel. Let η′i =

∧

j 6=i

ηj . For θi ∈ ConAi let θi = p−1i θi ∈ Con A. Note

g sends (θi)i∈I to λ∧ (∧

I θi). If θi, ψi ∈ ConAi, i ∈ I, we wish to show

[g((θi)i∈I), g((ψi)i∈I)] = g(([θi, ψi])i∈I).

Let α and β be the left and right sides of this equation. By Propo-sition 4.4(1) [θi, ψi] = p−1

i [θi, ψi] = [θi, ψi] ∨ ηi. Hence β = λ ∧


(∧

I([θi, ψi] ∨ ηi)) and α = [α ∧ (∧

θi), λ ∧ (∧

ψi)]. Now α ≤ β fol-lows from monotonicity.

To see the other inclusion first note that (by arguing with elementsin the direct product) we have λ ∧ (

∧

I θi) =∨

I θi ∧ η′i. Hence β =

∨

I [(θi ∧ ψi) ∨ ηi] ∧ η′i. Now θi = (θi ∧ η

′i) ∨ ηi by modularity. Hence

[θi, ψi] ∨ ηi = [(θi ∧ η′i) ∨ ηi, ψi ∧ η

′i] ∨ ηi = [θi ∧ η

′i, ψi ∧ η

′i] ∨ ηi by

additivity. Now this and modularity (and [θi ∧ η′i, ψi ∧ η

′i] ≤ η′i) yield

([θi, ψi] ∨ ηi) ∧ η′i = [θi ∧ η

′i, ψi ∧ η

′i]. Now, since η′i ≤ λ, β ≤ α by

monotonicity.The proof of (4) is similar to the proof of α ≤ β. �

Remarks 4.6. The homomorphism property, Proposition 4.4(1) orDefinition 3.1(1), is extremely useful in applications. Con B is natu-rally isomorphic with the interval 1/π of Con A under f. In the pres-ence of additivity, Proposition 4.4(1) simply tells us that for θ, ψ ∈ 1/πwe can compute the commutator of f(θ) and f(ψ) already in Con A.It corresponds to the congruence [θ, ψ]π = [θ, ψ] ∨ π. We call the oper-ation [θ, ψ]π a relative commutator.

If α/β ր γ/δ in Con A, then the map θ 7→ θ ∨ δ for θ ∈ α/β is alattice isomorphism with inverse ψ 7→ ψ∧α, ψ ∈ γ/δ. This is a classicalresult of modular lattice theory. The above isomorphism also respectsthe relative commutator. That is, if θ, ψ ∈ α/β and θ′, ψ′ ∈ γ/δ, then

[θ, ψ]β ∨ δ = [θ ∨ δ, ψ ∨ δ]δ

and

[θ′, ψ′]δ ∧ α = [α ∧ θ′, α ∧ ψ′]β.

The first equation is an easy consequence of additivity. To see thesecond, first note θ′ = (θ′ ∧ α) ∨ δ. Hence

[θ′, ψ′]δ ∧ α = ([θ′, ψ′] ∨ δ) ∧ α

= ([(θ′ ∧ α) ∨ δ, (ψ′ ∧ α) ∨ δ] ∨ δ) ∧ α

= ([θ′ ∧ α, ψ′ ∧ α] ∨ δ) ∧ α

= [θ′ ∧ α, ψ′ ∧ α] ∨ (δ ∧ α)

= [θ′ ∧ α, ψ′ ∧ α]β.

Our proof of (3) shows that g preserves any global congruence op-eration on V (which is modular) as long as it satisfies Definition 3.1(1)and is finitely additive.

We present now an example showing that the inequality in (4) maybe strict. Let Ai be the ring of all polynomials over the rationals, with0 constant term, in noncommuting indeterminates x1, ..., xi. Of course,


each congruence corresponds uniquely to a (two-sided) ideal and if Iand J are ideals [I.J ] = IJ ∨ JI where IJ is the ideal generated by{ab : a ∈ I, b ∈ J}. Clearly x21∨...∨x

2i ∈ [Ai,Ai]. Thus, if a ∈ A = ΠAi

has ith component x21+· · ·+x2i , then a ∈ Π[Ai,Ai]. However, a 6∈ [A,A].

If it were, then there would be elements a1, ..., ak, b1, ...bk in A witha = a1b1 + · · · + akbk. Choose n > k and let the nth component of aiand bi be pi and qi respectively. Then

(∗) x21 + · · ·+ x2n = p1q1 + · · ·+ pkqk.

Let pi be the polynomial obtained from pi by deleting all nonlinearterms and define qi similarly. Since the constant terms of each of thesepolynomials are zero, the above equation is still valid if we replace piand qi with pi and qi. Thus we may assume pi and qi are linear: say,pi =

∑nj=1 cijxj and qi =

∑nj=1 dijxj , cij , dij rational. Then (∗) yields

k∑

i=1

cisdit = δst 1 ≤ s, t ≤ n

i.e.,

CTD = In

where C = (cij), D = (dij) and CT is the transpose of C. Since C is a

k by n matrix over the rationals, with k < n, this is impossible.

We will now show that our commutator is identical with the onedefined by Hagemann and Herrmann, and also with CV defined in Def-inition 3.1. For the next few paragraphs θ and ψ will denote fixedcongruences of an algebra A. A(θ) will denote θ viewed as a subal-gebra of A2. Of course. A(ψ) is defined similarly. However, we shall

write the elements of A(θ) as

[

xy

]

instead of 〈x, y〉. We shall write the

elements of A(ψ) as[

x y]

instead of 〈x, y〉. Thus we are thinking ofthe elements of A(θ) as column vectors and those of A(ψ) as row vec-

tors. Hence, in the following definition, an element

[

a bc d

]

of A(θ, ψ),

can be thought of as a member of A(θ)×A(θ) and also ofA(ψ)×A(ψ).

Definition 4.7. (1) A(θ, ψ) ⊆ A4 is the subalgebra of allmatrices whose columns belong to θ and rows belong to ψ.

(2) ∆θ,ψ is the congruence onA(θ) generated by the set of all pairs

of the form

[

u vu v

]

in A(θ, ψ).


(3) ∆θ,ψ is the congruence on A(ψ) generated by the set of all

pairs of the form

[

x xy y

]

in A(θ, ψ).

In this notation,

[

x ry s

]

∈ ∆θ,ψ is equivalent to

[

xy

]

≡

[

rs

]

(mod ∆θ,ψ).

Another aspect of the symmetry of the commutator is that ∆θ,ψ = ∆θ,ψ.You will get a chance to prove this in the exercises.

Lemma 4.8. ∆θ,ψ is the least transitive relation on A(θ) containingM(θ, ψ).

Proof. It is not hard to show that the transitive closure ofM(θ, ψ),thought of as a relation on A(θ), is a congruence Γ on A(θ). Since thegenerators of ∆θ,ψ lie in M(θ, ψ),∆θ,ψ ⊆ Γ. But by Proposition 3.3(1)M(θ, ψ) ⊆ ∆θ,ψ. Hence. Γ ⊆ ∆θ,ψ. �

Theorem 4.9. For x, y ∈ A the following are equivalent:

(i) 〈x, y〉 ∈ [θ, ψ],

(ii)

[

x yy y

]

∈ ∆θ,ψ,

(iii) for some a,

[

x ay a

]

∈ ∆θ,ψ,

(iv) for some b,

[

x yb b

]

∈ ∆θ,ψ.

Proof. Clearly (ii) implies both (iii) and (iv). To see that (iii)

implies (ii), suppose

[

x ay a

]

∈ ∆θ,ψ. It is easy to see that this im-

plies that y ψ a. Therefore

[

a ya y

]

∈ ∆θ,ψ, and hence by transitivity,[

x yy y

]

∈ ∆θ,ψ, proving (ii).

Now assume that

[

x yb b

]

∈ ∆θ,ψ. Since the columns are in A(θ), x,

y, and b are all in the same θ–class. Thus. if we let η0, η1 ∈ ConA(θ)be the congruences associated with the two projections, we have thesituation diagrammed in Figure 1.

Hence,

[

x yy y

]

∈ ∆θ,ψ by the Shifting Lemma. Thus (ii), (iii), and

(iv) are all equivalent. We claim that [θ, ψ] is the least congruence


η0

η1 ∆θ,ψ

[

xy

]

[

yy

]

[

xb

]

[

yb

]

Figure 1.

on A which is a union of ∆θ,ψ classes. Indeed, by its definition (andProposition 4.2), [θ, ψ] is the smallest congruence β such that if onecolumn of a matrix in M(θ, ψ) is in β then the other is. Now if

[

ab

]

≡

[

cd

]

(mod ∆θ,ψ),

then, since ∆θ,ψ is the transitive closure of M(θ, ψ), there is a finitesequence of columns going from

[

ab

]

to

[

cd

]

with each consecutive pair of columns forming a matrix in M(θ, ψ). Itfollows from the above, that if

[

ab

]

∈ [θ, ψ], then

[

cd

]

∈ [θ, ψ],

i.e. [θ, ψ] is a union of ∆θ,ψ classes. Of course, any congruence whichis the union of ∆θ,ψ classes will satisfy the defining property of [θ, ψ].Thus the claim holds. From this, (ii) implies (i) follows immediately

since certainly

[

yy

]

∈ [θ, ψ].

Using that (ii) and (iv) are equivalent, it is not hard to show thatthe set of

[

xy

]

which satisfy (ii) is a congruence α.Now α is obviously the union of ∆θ,ψ

classes, and hence [θ, ψ] ≤ α, by the claim. Thus (i) implies (ii). �

The set of 〈x, y〉 which satisfy condition (iv) of the last theoremis the Hagemann-Herrmann commutator, and of course the theoremshows that it is the same as the one we have defined.

Theorem 4.10. C(V)(θ, ψ) = [θ, ψ].


Proof. By Proposition 4.4 [θ, ψ] ⊆ C(V)(θ, ψ). For the other inclu-sion we repeat the argument for groups given in Chapter 1. Considerthe following congruences on A(θ):

∆ = ∆θ,ψ

[θ, ψ]0 = {〈〈x, u〉, 〈y, v〉〉 ∈ A(θ)2 : 〈x, y〉 ∈ [θ, ψ]}

η1 = {〈〈x, u〉, 〈y, u〉〉 ∈ A(θ)2 : x θ y θ y}

and let P0 : A(θ) → A be the first projection. In Con A(θ) wehave C(V)(η1,∆) ≤ η1 ∧ ∆ ≤ [θ, ψ]0. The first inequality is part ofthe definition of C(V), and the second follows from the equivalence of(i) and (iv) in the last theorem. If for α ∈ ConA we put α0 = p−1

0 (α)and η0 = p−1

0 (0), then it is easy to see, by arguing on elements, thatθ0 = η1 ∨ η0, ψ0 = ∆ ∨ η0. Hence, θ = p0(η1 ∨ η0) and ψ = p0(∆ ∨ η0).Thus by the definition of C(V) we have

C(V)(θ, ψ) = C(V)(P0(η1 ∨ η0), P0(∆ ∨ η0))

= P0(C(V)(η1,∆) ∨ η0)

≤ [θ, ψ].

completing the proof. �

The next theorem gives the Hagemann-Herrmann lattice theoreticcharacterization of the commutator. First we introduce some usefulnotation. For B a subalgebra of A×A and γ ∈ ConA, let γi, i = 0, 1,be the congruence on B defined by 〈x0, x1〉 γi 〈y0, y1〉 if xi γ yi. We useη1 to denote 0i.

Theorem 4.11. Let A be an algebra in a modular variety V andlet α, θ, ψ ∈ ConA . Then a ≥ [θ, ψ] if and only if there is a B ∈ V

and a homomorphism f : B ։ A and σ, τ ∈ ConB with.

σ ∨ f−1(α) ≥ f−1(ψ)

τ ∨ f−1(α) ≥ f−1(θ)

f−1(α) ≥ σ ∧ τ.

Proof. First suppose that α ≥ [θ, ψ], and let β = [θ, ψ]. TakeB = A(θ), σ = ∆θ,ψ, τ = η1 ∈ ConB, and let f be the first projection.Then ∆θ,ψ ∨ β0 = ψ0, η1 ∨ β0 = θ0 = θ1, and β0 ≥ ∆θ,ψ ∧ η1. As notedin the last theorem, the last inequality follows from Theorem 4.9, andthe others follow from easy arguments on the elements. For example,if 〈x, y〉 ψ0 〈u, v〉, i.e., x π u and 〈x, y〉, 〈u, v〉 ∈ A(θ), then 〈x, y〉 β0〈x, x〉 ∆ 〈u, u〉 β0 〈u, v〉. Since β0 = f−1(β) ≤ f−1(α), one direction ofthe theorem is proved.

EXERCISES 33

For the other direction, suppose such a B, f, σ and τ exist. Thenby additivity.

[f−1(θ), f−1(ψ)] ≤ [τ ∨ f−1(α), σ ∨ f−1(α)]

≤ σ ∧ τ ∨ f−1(α)

= f−1(α).

Hence by Proposition 4.4, [θ, ψ] ≤ α, �

The next definition defines the residuation operation which is im-portant in multiplicative ideal theory. It also will play an importantrole here.

Definition 4.12. For θ, ψ ∈ ConA we let (θ :ψ) denote the largestγ ∈ ConA such that [ψ, γ] ≤ θ. This operation is known as residua-tion.

Exercises

1. Suppose that A lies in a permutable variety with Mal’cev termp(x, y, x). Suppose θ, ψ ∈ ConA with [θ, ψ] = 0 and let b ∈ Abe fixed. Define x+ y = p(x, b, y). Show that if x θ b ψ y andz is arbitrary then x+ y = y+x and x+(y+ z) = (z+ y)+ z.

2. Show that[

x yu v

]

∈ ∆θ,ψ if and only if

[

x uy v

]

∈ ∆ψ,θ.

3. Show that if ∆ is either ∆θ,ψ or ∆θ,ψ then(i)

[

x yx y

]

∈ ∆ if and only if 〈x, y〉 ∈ ψ.

(ii)[

x xu u

]

∈ ∆ if and only if 〈x, u〉 ∈ θ.

(iii)[

x yu v

]

∈ ∆

implies[

x yx y

]

,

[

x xu u

]

,

[

u vx y

]

,

[

y xv u

]

∈ ∆.


4. Use the Shifting lemma to show that if[

a ac d

]

∈ ∆θ,ψ

and[

ab

]

∈ θ,

then[

b bc d

]

∈ ∆θ,ψ.

5. Prove ∆θ,ψ = ∆θ,ψ. (Hint: By Lemma 4.8 it suffices to showthat if

[

x yu v

]

,

[

u vr s

]

∈ ∆θ,ψ

then[

x yr s

]

∈ ∆θ,ψ.

Do this using Lemma 2.3 and the previous exercises.)

6. For αi, β ∈ ConA show that the residuation defined in Defi-nition 4.12 satisfies (

∧

αi : β) =∧

(αi : β).

CHAPTER 5

The Fundamental Theorem on Abelian Algebras

Initially we do not assume A lies in a modular variety.

Definition 5.1. The center of any algebra A is the binary relationζA ⊆ A2 defined by

〈x, y〉 ∈ ζA ⇔ (∀t)(∀u, v) (t(u, x) = t(v, x)←→ t(u, y) = t(v, y)).

(The first quantifier is over all term operations of A, the second is overall n-tuples of A, n depending on t.)

Lemma 5.2. ζA is is the largest congruence α on A such that[α, 1A] = 0A. �

This result follows easily from Definition 3.2. Henceforth, all vari-eties will be assumed to be modular and all algebras will be assumedto generate modular varieties.

Definition 5.3. A congruence θ ∈ ConA is called Abelian if[θ, θ] = 0. An algebra A is Abelian if [1A, 1A] = 0A, or, what amountsto the same thing, ζA = 1A, variety is Abelian if all its members are.

Definition 5.4. We call A affine if there is an Abelian group〈A,+,−〉 = A having the same universe as A, and a 3-ary term oper-ation of A, t(x, y, z), such that:

(1) t(x, y, z) = x− y + z for all x, y, z ∈ A(2) f(x− y + z) = f(x)− f(y) + f(z) for each term operation f

and x,y, z ∈ An.

When these conditions hold, t is called a difference operation forA, the algebra 〈A, t〉 is called a ternary group, and we say that A ist-affine.

Before proceeding further, we mention that Definition 5.4(2) hastwo equivalent forms, as the reader can verify. First, {〈x, y, z, u〉 :x+y = z+u} is a subalgebra of A4. Second, each operation (and eachterm operation) of the algebra A is affine with respect to the group

A, that is to say, for any given operation F (say F is n-ary) there are

endomorphisms α1, . . . , αn of A and an element a ∈ A such that F can

35

36 5. THE FUNDAMENTAL THEOREM ON ABELIAN ALGEBRAS

be expressed identically as

F (x1, . . . , xn) =n

∑

i=1

αi(xi) + a.

In this chapter we shall prove Herrmann’s result that in a modularvariety every Abelian algebra is affine, and conversely. The proof wepresent is relatively short and direct. It also proves a generalizationand establishes other facts we will use later. The proof combines ideasfrom several people. Credits will be given at the end of the chapter.We begin with some useful preliminary results.

Theorem 5.5. For each modular variety V there is a ternary termd, called a difference term, satisfying the following.

(i) d(x, x, y) ≈ y is an identity of V.

(ii) If 〈x, y〉 ∈ θ ∈ ConA, A ∈ V, then d(x, y, y) [θ, θ] x.

(iii) If α, β, γ ∈ ConA, A ∈ V and α∧β ≤ γ, then the implicationof Figure 1 holds.

β

α γ

x

y

z

u′u

implies γ

x

y

z

u′ud(u, u′, y)

Figure 1.

Proof. Since V is modular there are termsm0(x, y, z, u), . . . , mn(x, y, z, u)which satisfy the identities of Theorem 2.2. Define qi(x, y, z), i =0, 1, . . . , n inductively by

q0(x, y, z) = z

qi+1(x, y, z) =

{

mi+1(qi(x, y, z), y, x, qi(x, y, z)) i odd

mi+1(qi(x, yz), x, qi(x, y, z)) i even

and set

d(x, y, z) = qn(x, y, z).

It follows from Theorem 2.2(ii) that qi(x, x, y) ≈ y.Hence d(x, x, y) ≈y holds in V.

5. THE FUNDAMENTAL THEOREM ON ABELIAN ALGEBRAS 37

To show (ii) assume 〈x, y〉 ∈ θ. We prove inductively that

(1)qi(x, y, y) [θ, θ] mi(y, y, y, x), i odd

qi(x, y, y) [θ, θ] mi(y, y, x, x), i even

Since mn(x, y, z, u) = u, (ii) will then follow. The case i = 0 is imme-diate from Day’s identities. Suppose i is odd and that (1) holds for i.Then

qi+1(x, y, y) = mi+1(qi(x, y, y), y, x, qi(x, y, y))

[θ, θ] mi+1(mi(y, y, y, x), y, x,mi(y, y, y, x)).

Now by Theorem 2.2

mi+1(mi(y, y, y, x),y, y,mi(y, y, y, x))

= mi(y, y, y, x)

= mi+1(y, y, y, x)

= mi+1(mi(y, y, y, y), y, y,mi(x, x, x, x).

The term condition of Definition 3.2(2) says that if

t(x, x1, . . . , xn−1) = t(x, y1, . . . , yn−1)

for certain elements ofA with xi θ yi, i = 1, . . . , n−1, then t(y, x1, . . . , xn−1) [θ, θ]t(y, y1, . . . , yn−1) for any y ∈ A with x θ y. Applying this to the sixthvariable of the above equation, we get

mi+1(mi(y, y, y, x), y, x,mi(y, y, y, x))

[θ, θ] mi+1(mi(y, y, y, y), y, x,mi(x, x, x, x)) = mi+1(y, y, x, x).

Thus qi+1(x, y, y) [θ, θ] mi+1(y, y, x, x).Now suppose i is even and that (1) holds for i. Then

qi+1(x, y, y) [θ, θ] mi+1(mi(y, y, x, x), x, y,mi(y, y, x, x))

as before. Now

mi+1(mi(y, y, x, x),x, x,mi(y, y, x, x))

= mi(y, y, x, x)

= mi+1(y, y, x, x)

= mi+1(mi(y, y, y, y), y, x,mi(x, x, x, x)).

The term condition implies

mi+1(mi(y, y, x, x),x, y,mi(mi(y, y, x, x))

[θ, θ] mi+1(mi(y, y, y, y), y, y,mi(x, x, x, x))

= mi+1(y, y, y, x).


Thus qi+1(x, y, y) [θ, θ] mi+1(y, y, y, x).To prove (iii) we shall show that any term d(x, y, z) which satisfies

(i) and (ii) satisfies (iii). Suppose d(x, y, z) satisfies (i) and (ii), andthat the relations of Figure 2 hold for congruences α, β, and γ whereα ∧ β ≤ γ.

β

α γ

x

y

z

u′u

Figure 2.

Then d(u, u′, y) β d(y, y, y) = y. Also

d(u, u′, y) γ d(z, u′, y)(2)

d(z, u′, y) α d(z, z, x) = x(3)

d(z, u′, y) β d(x, y, y).(4)

Since 〈x, y〉 ∈ α∧ (β ∨ γ), (ii) yields d(x, y, y) [α ∧ (β ∨ γ), α ∧ (β ∨ γ)]x. But [α ∧ (β ∨ γ), α ∧ (β ∨ γ)] ≤ [α, β ∨ γ] = [α, β] ∨ [α, γ] ≤(α ∧ β) ∨ (α ∧ γ) = α ∧ γ. Thus by (4) d(z, u′, y) β ∨ α ∧ γ x. Henced(z, u′, y) α ∧ (β ∨ (α ∧ γ)) x. But α ∧ (β ∨ (α ∧ γ)) = (α ∧ β) ∨ (α ∧γ) = α ∧ γ, that is, d(z, u′, y) α ∧ γ x. Hence by (2) d(u, u′, y) γ x,proving (iii). �

We remark that a term which satisfies (iii) will satisfy (i) and (ii);see Exercise 6.

The term d(x, y, z) constructed above plays an important role inour theory. We call a term which satisfies Theorem 5.5(i) and (ii) adifference term for V , or a Gumm difference term. In what fol-lows, d will always denote a difference term. Of course, in a permutablevariety a Mal’cev term satisfies Theorem 5.5(i) and (ii) and thus is adifference term.

The next lemma proves Gumm’s observation that a Mal’cev term(see Theorem 2.1(1)) which commutes with itself defines a ternaryAbelian group.

Lemma 5.6. Let t(x, y, z) be a 3-ary operation on a set S such that tcommutes with itself and satisfies the Mal’cev equations t(x, x, y) ≈ y ≈t(y, x, x, ). For any fixed a ∈ S the operation x + y = t(x, a, y) defines


an Abelian group on S with a as null element and with −x = t(a, x, a).Moreover, t(x, y, z) = x− y + z.

Proof. That t commutes with itself means that for all xi, yi, zi ∈ S

t(t(x1, y1, z1),t(x2, y2, z2), t(x3, y3, z3))

= t(t(x1, x2, x3), t(y1, y2, y3), t(z1, z2, z3)).

Clearly, x+ a = a + x = x. For associativity

x+ (y + z) = t(x, a, t(y, a, z))

= t(t(x, a, a), t(a, a, a), t(y, a, z))

= t(t(x, a, y), t(a, a, a), t(a, a, z))

= t(t(x, a, y), a, z)

= (x+ y) + z.

To see that −x = t(a, x, a), we calculate

x+ t(a, x, a) = t(x, a, t(a, x, a))

= t(t(x, a, a), t(a, a, a), t(a, x, a))

= t(t(x, a, a), t(a, a, x), t(a, a, a))

= t(x, x, a)

= a.

Now

x+ y = t(x, a, y)

= t(t(a, a, x), t(a, a, a), t(y, a, a))

= t(t(a, a, y), t(a, a, a), t(x, a, a))

= t(y, a, x)

= y + x.

To see that x− y+ z = t(x, y, z), we first show that x− y = t(x, y, a) :

x− y = t(x, a, t(a, y, a))

= t(t(x, a, a), t(a, a, a), t(a, y, a))

= t(t(x, a, a), t(a, a, y), t(a, a, a))

= t(x, y, a).

Finally

x− y + z = t(t(x, y, a), a, z)

= t(t(x, y, a), t(a, a, a), t(a, a, z))

= t(t(x, a, a), t(y, a, a), t(a, a, z))t(x, y, z).


�

Proposition 5.7. Let α ≥ β where α, β ∈ ConA. The followingare necessary and sufficient conditions in order that [α, β] = 0. Forany basic operation (and hence any term operation) s(x1, . . . , xn) andelements xi β yi α zi, (i = 1, . . . , n) we have

d(s(x), s(y), s(z)) = s(d(x1, y1, z1), . . . , d(xn, yn, zn))

and

〈y, z〉 ∈ β implies d(y, z, z) = d(z, z, y) = y.

Proof. Suppose first that [α, β] = 0. By considering the congru-ences η0, η1 (first and second projections) and ∆ = ∆α,β on the algebraA(α) (see Definition 4.7). When x β y α z , we have the relations ofFirure 3.

η1

η0 ∆

〈z, y〉

〈z, x〉

〈y, y〉

〈y, x〉〈x, x〉

Figure 3.

Hence, by Theorem 5.5(iii)

〈d(x, y, z), x〉 ≡ 〈z, y〉 (mod ∆α,β).

In particular

〈d(s(x), s(y), s(z)), s(x)〉 ≡ 〈s(z), s(y)〉 (mod ∆α,β).

But applying s to the n congruences 〈d(xi, yi, zi), xi〉 ≡ 〈zi, yi〉 yields

〈s(d(x1, y1, z1), . . . , d(xn, yn, zn)), s(x)〉 ∆α,β 〈s(z), s(y)〉.

Since [α, β] = 0, and by the equivalence of (i) and (iv) in Theorem 4.9the two displayed formulas above give the desired conclusion.

The last statement follows from Theorem 5.5.Now suppose the conditions hold. Then the congruence ∆β,α on

A(β) (see Definition 4.7) is characterized by

〈x, y〉 ∆β,α 〈u, v〉 iff x β y α u and v = d(y, x, u).

To see this let ∆′ be the relation on A2 defined by the second part,that is, ∆′ = {〈〈x, y〉, 〈u, v〉〉 : x β y α u, v = d(y, x, u)}. That ∆′

determines a congruence on A(β) follows from the properties of d andthe conditions of Proposition 5.7. For example, to see that this relation


is symmetric suppose x β y α u and v = d(y, x, u). Clearly u β v α xand since d commutes with itself.

d(v, u, x) = d(d(y, x, u), d(x, x, u), d(x, x, x))

= d(d(y, x, x), d(x, x, x), d(u, u, x)

= d(y, x, x) = y.

which proves 〈u, v〉 ∆′ 〈x, y〉, i.e., ∆′ is symmetric.Now ∆β,α is the congruence on A(β) generated by {〈〈x, x〉, 〈u, u〉〉 :

x α u}. Clearly these generators are in ∆′. Hence ∆β,α ⊆ ∆′. Sup-pose 〈x, y〉 ∆′ 〈u, v〉. Then x α u so that 〈x, x〉 ∆β,α 〈u, u〉. Now bycalculations in A(β) we have

〈x, y〉 = 〈d(x, x, x), d(y, x, x)〉 = d(〈x, y〉, 〈x, x〉, 〈x, x〉)

∆β,α d(〈x, y〉, 〈x, x〉, 〈u, u〉) = 〈d(x, x, u), d(y, x, u)〉 = 〈u, v〉.

Hence ∆β,α = ∆′.Now if x [α, β] y then 〈x, x〉 ∆β,α 〈x, y〉. Hence y = d(x, x, x), i.e.,

y = x. �

The next corollary reformulates Proposition 5.7 for the case α = β.For u ∈ A we let M(β, u) = 〈u/β, d〉, the ternary group with universethe β− block containing u. In Chapter 9 M(β, u) will be given amodule structure.

Corollary 5.8. A congruence β on A is Abelian if and only ifeach M(β, u) is a ternary group and whenever s(u1, . . . , un) = u wheres is an n–ary term operation then s : M(β, u1) × · · · ×M(β, un) →M(β, u) is a homomorphism, i.e., s is affine between the β–blocks.

Proof. Since d is idempotent, the β− block u/β is closed under d.If β is Abelian then, by Proposition 5.7, d commutes with itself on u/βand thus, by Lemma 5.6, M(β, u) is a ternary group. This corollaryfollows easily from Proposition 5.7. �

Corollary 5.9. In a modular variety every Abelian algebra isaffine, and conversely. �

As the definition of affine and the remarks following it indicate,Abelian algebras are closely related to modules (in fact, each Abelianalgebra is polynomially equivalent to a module). This connection isstudied more thoroughly in Chapter 9.

The term d(x, y, z) constructed in Theorem 5.5 was first constructedby Herrmann in [45]. The short proof that it satisfies (ii) is Taylor’s.Gumm [36], [38], using his geometrical methods, constructed a termsatisfying (iii) and showed (iii) implies (i) and (ii). Udi Hrushovskii


proved that (i) and (ii) imply (iii). Proposition 5.7 was proved byGumm in [38]. Corollary 5.8 appeared in Herrmann [45]. Of coursethe fundamental theorem on Abelian algebras, Corollary 5.9, is Her-rmann’s [45]. It had been previously established for permutable vari-eties independently by Gumm and McKenzie.

Proposition 5.7 will play an important role in the rest of this book.Kiss [56] obtained a nice generalization of Theorem 5.5 and Proposi-tion 5.7. He constructed a 4–ary term for each modular variety whichhe calls a 4–difference term. He then gave necessary and sufficient con-ditions for [α, β] = 0 similar to those of Proposition 5.7. However histheorem does not require that α ≥ β. His results are described in moredetail in the Related Literature chapter.

EXERCISES 43

Exercises

All algebras are assumed to belong to a modular variety.

1. Show that if M3 (the five element modular, nondistributivelattice) is a 0, 1–sublattice of Con A then A is Abelian.(Assume V (A) is modular.)

2. Prove that for an algebra A in a modular variety the followingare equivalent.(i) M3 is a 0, 1–sublattice of Con(A×A) ,(ii) the projection congruences of A×A have a common com-

plement,(iii) M3 is a 0, 1–sublattice of some subdirect product of two

copies of A.(iv) A is Abelian.

3. Let B be a subdirect power of two copies of A such that thejoin of the kernels of the two projections is 1. Show that if Bcontains M3 as a 0, 1–sublattice then the subdirect product isdirect.

4. Let α and β be the projection congruences of A × B and letγ ∈ Con(A×B) be arbitrary. Use Theorem Theorem 5.5(iii)to prove Gumm’s result that every congruence permutes withthe projection congruences of a direct product.

5. Suppose α ≥ β and [α, β] = 0. Then for 〈u, v〉 ∈ α prove thatthe function x 7→ d(x, u, v) is an isomorphism from M(β, u)onto M(β, v).

6. Let V be a variety which has a term d(x, y, z) satisfying Theo-rem 5.5(iii). Show that V is modular and Theorem 5.5(i) and(ii) hold for this d.

7. Show that if 〈A, ·, /, \, 1〉 is a loop which is Abelian then x · yis commutative and associative, i.e., x · y is an Abelian groupoperation.


8. Let G be the quasigroup with multiplication table

· 0 1 2 30 3 2 0 11 2 3 1 02 1 0 2 33 0 1 3 2

Show G is Abelian (but x · y is neither commutative nor asso-ciative).

9. Let 〈A, ·, /, \〉 be the quasigroup with multiplication

· 0 1 2 30 0 1 2 31 2 3 0 12 1 2 3 03 3 0 1 2

For c ∈ A, define x+c y = p(x, c, y) where p(x, y, z) = (x/(y \y)) · (y \ z). Show that for each c, 〈A,+c〉 is an Abelian groupbut that 〈A, ·, /, \〉 is not Abelian.

10. A subloop H of a loop G is called a normal subloop if thereis some congruence on G whose class containing the identityelement is H. Exactly as for groups there is a one-to-one cor-respondence between normal subloops and congruences. Thusif H and K are normal subloops we can define [H,K] to be thenormal subloop corresponding to the commutator of the con-gruences associated with H and K. This exercise will showthat this correspondence is not as nice as one might hope.Namely if G is a group and H is a normal subgroup of G then[H,H] is the derived subgroup of H. In particular it dependsonly on H and not on G. However for loops this is not thecase; [H,H] is not determined from just H. One needs thewhole congruence associated with H to determine [H,H].

To see this let Z4 = {0, 1, 2, 3} denote the group of integersunder addition modulo 4. Let G = Z4×Z4 and define a binaryoperation on G by 〈a, b〉 · 〈c, d〉 = 〈a+c, b+d〉 unless b = d = 1

EXERCISES 45

in which case the operation is defined by the following table:

· 01 11 21 3101 12 02 22 3211 02 22 32 1221 22 32 12 0231 32 12 02 22

Show that G = 〈G, ·〉 is a loop with identity element 〈0, 0〉and that the second projection is a homomorphism. Let θbe the congruence associated with this projection. Let H ={〈0, 0〉, 〈1, 0〉, 〈2, 0〉, 〈3, 0〉} be the normal subloop associatedwith θ. Show that as a subloop of G, H is isomorphic to thegroup Z4, and hence H is an Abelian loop. However, showthat in G, [θ, θ] 6= 0.

This fact may explain why the commutator was slow to bedeveloped in the theory of loops.

11. Prove the following result of C. Herrmann [46]. Let V be avariety such that Con A is a complemented modular latticefor all A ∈ V. Prove that V is Abelian. The congruence λdefined in the proof of Proposition 4.5 may be useful.

CHAPTER 6

Permutability and a Characterization of

Modular Varieties

One of the surprising consequences of commutator theory is thatcertain pairs of congruences are forced to permute. Most of the peoplewho have worked with the commutator have proved results of this na-ture. For example, by Corollary 5.9 Abelian algebras have permutablecongruences and the factor congruences of the direct product of twoalgebras permute with all congruences (see Exercise 5.4.). The bestresults on permutability have been obtained by H.-P. Gumm [41]. Wepresent here a sample of these results, proving those parts we will uselater and relegating the proofs of the remaining results to the exer-cises. We also give Gumm’s characterization of modular varieties interms of a Mal’cev condition which amalgamates Jonsson’s conditionfor distributivity with Mal’cev’s condition for permutability (see The-orem 2.1).

Definition 6.1. Given α, β ∈ Con A we put [α]0 = (β, α]0 = α,[α]n+1 = [[α]n, [αn]], (β, α]n+1 = [β, (β, α]n] for each n < ω. Then

(1) α is Abelian if [α, α] = 0, central if [1, α] = 0. k-step solv-able if [α]k = 0. k-step nilpotent if (α, α]k = 0

(2) A is k-step solvable if 1A is k-step solvable, k-step nilpotentor nilpotent of class k if (1A, 1A]

k = 0.(3) α is co-solvable if A/α is solvable (i.e., k-step solvable for

some k).

In the following, α ◦β is the composition (or relation product) of αwith β.

Theorem 6.2. For α, β ∈ Con A we have for all k < ω

α ◦ β ⊆ [α]k ◦ β ◦ α.

Every solvable and every co-solvable congruence permutes with all con-gruences, and every solvable algebra has permuting congruences.

Proof. Induct on k. The result is trivial for k = 0. Supposex α y β z. Then by induction there are elements u, v ∈ A withx [α]k u β v α z. Then, by Theorem 5.5, x [α]k+1 d(x, u, u) β

47

48 6. PERMUTABILITY

d(x, u, v) α d(u, u, z) = z, since [α]k ≤ α. The statement about solv-able congruences follows immediately. If α is co-solvable then [1]k ≤ αfor some, k. If β ∈ Con A is arbitrary, then [β]k ≤ [1]k ≤ α. Henceβ ◦ α ⊆ [β]k ◦ α ◦ β = α ◦ β. �

Theorem 6.3. The following conditions are equivalent where α, β ∈Con A.

(i) α permutes with β(ii) [α]m permutes with [β]n for all m,n ∈ ω(iii) [α]m permutes with [β]n for some m,n ∈ ω.

This theorem is due to Gumm. His proof is outlined in the exercises.He obtains various stronger results in [1983].

Theorem 6.4. (Gumm [1981], Theorem 1.4) Let V be a variety (orquasivariety). Then V is modular if and only if there exist terms p andq0, . . . , qn for some n such that the following are identities of V :

(1) q0(x, y, z) ≈ x,(2) qi(x, y, x) ≈ x i ≤ n,(3) qi(x, y, y) ≈ qi+1(x, y, y) for i even,(4) qi(x, x, y) ≈ qi+1(x, x, y) for i odd,(5) qn(x, y, y) ≈ p(x, y, y),(6) p(x, x, y) ≈ y.

Proof. First suppose that V is modular and let A ∈ V, and letα, β, γ ∈ Con A with x α y β z and x γ z. Let θ = Cg(x, z) andnote that θ ≤ γ ∧ (α ∨ β). Hence [θ, θ] ≤ [γ, α ∨ β] = [γ, α] ∨ [γ, β] ≤(γ ∧ α) ∨ (γ ∧ β). Thus by Theorem 5.5 x (γ ∧ α) ∨ (γ ∧ β) d(x, z, z).So

(7) x (γ ∧ α) ∨ (γ ∧ β) d(x, z, z) β d(x, y, z) α d(x, x, z) = z.

Notice that this shows that every modular variety satisfies the congru-ence identity (α◦β)∧γ ≤ ((γ∧α)∨(γ∧β))◦β◦α. Now we apply this toF = FV(x, y, z) with α = Cg(x, y), β = Cg(y, z), and γ = Cg(x, z). By(7) there are terms q0(x, y, z) = x, q1(x, y, z), . . . , qn(x, y, z) = d(x, z, z)such that, in F, qi(x, y, z) β qi+1(x, y, z) if i is even, qi(x, y, z) αqi+1(x, y, z) if i is odd, and qi(x, y, z) γ qi+1(x, y, z) for i = 0, . . . , n.As in the proof of Theorem 2.1, β restricted to the subalgebra gener-ated by x and y is trivial, and thus for i even qi(x, y, y) β qi(x, y, z) βqi+1(x, y, z) β qi+1(x, y, y) implies that, in F, qi(x, y, y) = qi+1(x, y, y)and hence (3) holds. Similarly (2)–(4) hold. Now if we let p(x, y, z) =d(x, y, z) then (1), (5), and (6) hold as well.

EXERCISES 49

Now suppose that V has terms satisfying the equations of the the-orem. We may assume that n is even since if it is odd then equa-tions (3) and (5) imply that qn−1(x, y, y) ≈ p(x, y, y) and thus we couldomit qn. Define termsm0(x, y, z, u), . . . , m2n+2(x, y, z, u) as follows. Setm0(x, y, z, u) = x. If i is even (and 1 ≤ i ≤ n), set

m2i−1(x, y, z, u) = qi(x, y, u)

m2i(x, y, z, u) = qi(x, z, u).

If i is odd set

m2i−1(x, y, z, u) = qi(x, z, u)

m2i(x, y, z, u) = qi(x, y, u).

Finally let m2n+1(x, y, z, u) = p(y, z, u) and let m2n+2(x, y, z, u) = u.Then it is not difficult to show that these are Day terms for V, i.e.,they satisfy the equations of Theorem 2.2. �

The proof of the above theorem is based on Lakser, Taylor andTschantz [59]. The congruence identity used in the proof is due toSteve Tschantz and is actually equivalent to modularity (see the nexttheorem). Gumm’s original proof produced a similar congruence im-plication.

Theorem 6.5. A variety V is modular if and only if for all A ∈ V

and all α, β and γ ∈ Con A,

(8) (α ◦ β) ∧ γ ≤ (γ ∧ α + γ ∧ β) ◦ β ◦ α

Proof. The proof of Theorem 6.4 showed that (8) holds if V ismodular. If (8) holds let F = FV(x, y, z), and let α = Cg(x, y), β =Cg(y, z), and γ = Cg(x, z) in Con F. Since 〈x, z〉 is in the left side of(8) it is in the right. Now an argument similar to the one in Theorem 6.4yields terms q0, . . . , qn, p which satisfy Theorem 6.4(1)–(6). Hence V ismodular. �

Notice that our proof of Theorem 6.4 shows that any Gumm dif-ference term can serve as the p(x, y, z) of that theorem. One of theexercises proves the converse.

Exercises

1. Prove for α, β ∈ Con A, k < ω that

α ◦ β ⊆ β ◦ α ◦ [β]k.

(this can be derived directly from the inclusion of Theorem6.2. by reversing the roles of α and β.)

50 6. PERMUTABILITY

2. Use Theorem 6.2. to prove for all m,m ∈ ω

α ◦ β ⊆ β ◦ [α]m ◦ [β]n ◦ α.

Deduce (iii) → (i) in Theorem 6.3.

3. Prove (i) → (ii) in Theorem 6.3. First show that it sufficesto prove that if α, β ∈ Con A and α permutes with β then[α, α] permutes with β. Suppose x [α, α] y β z. Then thereis a u ∈ A with x β u α z. Use the Shifting Lemma to show〈u, z〉 ∈ [α, α]∨ (α∧β). Now use the previous exercise to show[α, α] and α ∧ β permute.

4. Show that if α and β are nilpotent congruences, then α ∨ βis nilpotent. Conclude that if Con A satisfies the ascendingchain condition then A has a unique largest nilpotent congru-ence.

5. Let V be a modular variety an let q0, . . . , qn, p be terms of Vwhich satisfy Theorem 6.4(1)–(6). Show that p is a Gummdifference term for V, i.e., show that for any x, y ∈ A ∈ V,x [θ, θ] p(x, y, y) if x θ y.

6. Let V be a modular variety with Gumm terms q0, . . . , qn, p (seeTheorem 6.4.). Suppose that A ∈ V and that a, b, c, d ∈ A.We write C(a, b, c, d), and say that 〈a, b〉 centralizes 〈c, d〉, ifand only if [CgA(a, b),CgA(c, d)] = 0A. This exercise involvesfinding an equational characterization of this 4–ary relation,equivalently, finding a uniformly describable set of generatorsfor the congruence [CgA(a, b),CgA(c, d)]. We begin by definingtwo other 4–ary relations.

H(a, b, c, d)⇐⇒

s(a, c, e) = s(a, d, e)→ s(b, c, e) = s(b, d, e)

for every term operation s(x, y, z) and for all e.

K(a, b, c, d)⇐⇒

qi(s(a, e),t(c, f), s(b, e)) = qi(s(a, e), t(d, f), s(b, e)) and

p(r(a, c, e), r(b, c, e), r(b, d, e))

= p(r(a, d, e), r(b, d, e), r(b, d, e)).

for all term operations s(x,y), t(x,y), and r(x, y, z), for everye and f , and for every i ≤ n. Using Gumm’s equations, derivethese facts:

EXERCISES 51

(1) C(a, b, c, d)→ H(a, b, c, d).

(2) H(a, b, c, d)→ K(a, b, c, d).

(3) K(a, b, c, d)→ H(c, d, a, b).

(4) H(a, b, c, d)←→ K(a, b, c.d).

(5) H(a, b, c, d)←→ H(a, b, d, c).

(6) H(a, a, c, d).

(7) H(a, b, c, d) and H(b, e, c, d)→ H(a, e, c, d).

(8) If f is a basic n-ary operation of A and if H(ai, bi, c, d)holds for each i < n, then H(u, v, c, d) holds, where u =f(a) and v = f(b).

(9) For each 〈a, b〉 the set K(a, b) = {〈c, d〉 : K(a, b, c, d)} isa congruence equal to (0A : Cg(a, b)).

Conclude that C(a, b, c, d)←→ K(a, b, c, d).

7. The starting assumptions are the same as in the previous ex-ercise. Now suppose that α and β are congruences of A andthat either α is Abelian, or that p satisfies the other Mal’cevequation p(x, y, y) ≈ x. Prove that [α, β] = 0A if and onlyif p : D → A is a homomorphism, where D is the subalge-bra of A3 consisting of the elements 〈x, y, z〉 with x α y β z.See Gumm’s characterization of Abelian congruences, Theo-rem 5.7.

8. Let A and B be algebras in a modular variety and supposethat the congruences of A permute, and the congruences of Bpermute. Prove that the congruences of A×B permute.

9. Let d(x, y, z) be a Gumm difference term for a modular varietyV and define d1(x, y, z) = d(x, y, z) and

dn+1(x, y, z) = dn(x, dn(x, y, y), dn(x, y, z))

Show that dn is also a difference term. Moveover, if 〈x, y〉 ∈θ ∈ ConA, A ∈ V, then dn(x, y, y) [θ]

n x.

CHAPTER 7

The Center and Nilpotent Algebras

In this chapter we show how an algebra A may be decomposed interms of an Abelian algebra and A/ζ where ζ is the center of A. Someeasy applications of this decomposition to nilpotent loops are given inthe exercises. We also investigate nilpotent algebras, showing that theyhave uniform and regular congruences and that associated with eachnilpotent algebra is a nilpotent loop. Finally we prove a propositionwhich played an important role in Burris-McKenzie [8].

Corollary 5.9 gave the structure of Abelian algebras. We shall nowdescribe the structure of an algebra over its center. Suppose that Q,B ∈ V, Q is Abelian with associated group Q = 〈Q,+,−〉 (so thatd(x, y, z) = x − y + z for x, y, z ∈ Q) and that we are given for eachbasic operation symbol Fi of V a map Ti : B

n :→ Q where Fi is n–ary.Write T = (Ti, i ∈ I) for the system of maps, and define an algebraA = Q⊗T B by taking Q×B for the universe of A and putting

FAi (〈q1, b1〉, . . . , 〈qn, bn〉) = 〈F

Qi (q) + Ti(b), F

(B)i (b)〉

for i ∈ I. Notice that since Q is Abelian and hence affine each termF for V has a corresponding transfer function T so that the aboveequation holds. This algebra Q⊗T B need not belong to V, of course,but we have the following.

Proposition 7.1. Given A ∈ V, let B = A/ζA. There is anAbelian algebra Q ∈ V and α system T as above so that A ∼= Q⊗T B,and furthermore

(1) the center of Q⊗T B is the kernel of the projection onto B;(2) Q = A(ζA)/∆ζA,1A.

Proof. Recall from Chapter 4 and Definition 5.1 thatA(ζA) is thecenter ofA (a congruence ) regarded as an algebra, and that congruence∆ = ∆ζA ,1A on this algebra is generated by {〈〈x, x〉, 〈y, y〉〉 : x, y ∈ A},As in the proof of Proposition 5.7 we have

(1) 〈x, y〉 ≡ 〈u, v〉 (mod ∆) ⇔ 〈x, y〉 ∈ ζ and v = d(y, x, y).

That Q is Abelian is proved as follows. Where η0, η1 are the pro-jection kernels on A(ζ), in Con A(ζ), we have η0 ∨ ∆ = η1 ∨ ∆ = 1

53

54 7. THE CENTER AND NILPOTENT ALGEBRAS

obviously. Thus [1, 1] ⊆ ∆ ∨ (η0 ∧ η1) = ∆ by additivity. Then inCon Q, [1, 1] = 0 by the homomorphism property of commutator (i.e.Proposition 4.4(1)).

Now let r ∈ AA be a choice function for the ζ–blocks, so that r(x)ζxand x ζ y if and only if f(x) = r(y). Then we get a one-one map of Aonto Q×B by defining

π(x) = 〈〈r(x), x〉/∆, x/ζ〉.

(That π is bijective follows from (1) above and Theorem 5.5). For F1

a basic operation of V and u ∈ Bn, say uj = xj/ζ , put

Ti(u1, . . . , un) = 〈r(Fi(x)), Fi(r(x1), . . . , r(xn))〉/∆.

We can choose 0 = 〈x, x〉/∆ as the identity element of Q; then we havea+ b = d(a, 0, b) for a, b ∈ Q.

What remains is to show that for Fi, u, x as above with say π(xj) =〈aj, uj〉 for 1 ≤ j ≤ n, that we have

π(Fi(x)) = 〈Fi(a) + Ti(u), Fi(u)〉.

This is equivalent to showing

〈r(Fi(x)), Fi(x)〉/∆ = 〈Fi(rx), Fi(x)〉/∆+ 〈r(Fi(x)), Fi(rx)〉/∆.

If we use that 0 = 〈Fi(rx), Fi(rx)〉/∆ and a + b = d(a, 0, b), and thatd(y, z, z) = y when y ζ z, then the computation checks out. �

Corollary 7.2. An algebra in a modular variety V is 2–step nilpo-tent if and only if it can be represented as Q0⊗

T Q1 where Q0 and Q1

are Abelian algebras in V.

Proof. On one hand, if A in Proposition 7.1 is 2–step nilpotent,then B is Abelian. On the other hand, if A = Q0 ⊗

T Q1 ∈ V andQ0,Q1 ∈ V are Abelian, then it is clear (from Def. 5.1.) that theprojection congruence η of A onto Q1 satisfies η ≤ ζA, and true alsothat [1, 1] ≤ η since Q1 is Abelian. �

The next few results give additional information about nilpotentalgebras. They show nilpotency has some rather strong consequences.Exercise 7, at the end of the chapter, shows that solvable algebras donot have these properties in general. By Proposition 4.4 and Propo-sition 4.5 the nilpotent algebras of class k in a modular variety froma subvariety. By Theorem 6.2, if A is nilpotent then V (A) is per-mutable. Thus for the next few results we assume our algebras liein a permutable variety U with Mal’cev term p(x, y, z). Define termsfn(y, b, c) inductively by f0(y, b, c) = y and

fn+1(y, b, c) = p(b, p(b, y, p(fn(y, b, c), b, c)), fn(y, b, c)).

7. THE CENTER AND NILPOTENT ALGEBRAS 55

The congruence (1, 1]k defined in Definition 6.1. will be denoted(1]k. A is nilpotent if only if (1]k = 0 for some k. The next lemmashows that the function x 7→ fn(x, b, c) is the inverse to the functionx 7→ p(x, b, c) if A is nilpotent of class n.

Lemma 7.3. If A ∈ U and x, y, b, c ∈ A the for all n

fn(p(x, b, c), b, c) (1]n x.

andp(fn(y, b, c), b, c) (1]n y.

Proof. We prove the second relation: the proof of the first issimilar but easier. Induct on n. Since (1]0 = 1 the result is trivial forn = 0. Let y′ = p(fn(y, b, c), b, c). By induction y′(1]ny. In the algebraA/(1]n+1, (1]n is contained in the center. Hence by Theorem 5.7 (sinceb = p(b, y, y) (1]n p(b, y, y

′))

p(fn+1(y, b, c), b, c) = p((p(b, p(b, y, y′), fn(y, b, c)), b, c)

= p(p(b, p(b, y, y′), fn(y, b, c)), p(b, b, b), p(b, b, c))

≡ p(p(b, b, b), p((p(b, y, y′), b, b), p(fn(y, b, c), b, c)) (mod (1]n+1)

= p(b, p(b, y, y′), y′).

Thus p(fn+1(y, b, c), b, c) (1]n+1 p((b, p(b, y, y′), y′). The following ma-

trix lies in M((1]n, 1) (see Definition 3.2).[

p(b, p(b, y, y′), y′) p(b, p(b, b, y′), y′)p(b, p(b, y, y), y) p(b, p(b, b, y), y)

]

=

[

p(b, p(b, y, y′), y′) by b

]

Hence

y (1]n+1 p(b, p(b, y, y′), y′) (1]n+1 p(fn+1(y, b, c), b, c).

�

Corollary 7.4. If A ∈ U is nilpotent and b, c ∈ A then the func-tion x 7→ p(x, b, c) is on-one and onto. Moreover if θ ∈ Con A thenthis function, restricted to b/θ, is a bijection from b/θ to c/θ.

Proof. The first statement is clear from Lemma 7.3. If f denotesthe restricted function then f is clearly one-one and if x ∈ b/θ thenf(x) ∈ c/θ. Since A/θ is also nilpotent the function, x 7→ p(x, b, c)induces a permutation of the θ–blocks. Hence the inverse images ofevery element of c/θ under this function lie in b/θ. Since x 7→ p(x, b, c)is onto it follows that f is also. �

Corollary 7.5. If A is nilpotent then A has uniform congruences(i.e., all the blocks of any congruence θ have the same size). �


Lemma 7.6. If A ∈ U is nilpotent and a, b, c ∈ A then

Cg(a, b) = Cg(p(a, b, c), c)

Proof. Suppose (1]n = 0 in A. Then p(fn(c, b, c), b, c) = c =p(b, b, c). Hence fn(c, b, c) = b. Now let ψ = Cg(p(a, b, c), c). Clearlyψ ≤ Cg(a, b). But p(a, b, c) ψ c implies a = fn(p(a, b, c), b, c) ψfn(c, b, c) = b. �

Corollary 7.7. If A is nilpotent then A has regular congruences,i.e., if two congruences have a block in common, they are equal. �

In Chapter 5 we saw that Abelian algebras were closely connectedwith Abelian groups. Nilpotent algebras have a close connection withnilpotent loops. For suppose A ∈ U is nilpotent. Let 0 be an arbitraryelement of A and define x+ y = p(x, 0, y). Then by Corollary 7.4. thisdefines a loop with null element 0. Moreover by Lemma 7.3. the leftand right division operations (perhaps we should call them subtractionoperations) are also polynomials on A. Also note that by Theorem 5.7if a ζ 0 and x, y ∈ A then a + x = x + a, a + (x + y) = (a + x) + y,and (x+ a) + y = x+ (a+ y).

In Exercise 7 we show how to construct solvable algebras in whichthe properties above fail.

Proposition 7.8. Let B be subdirectly irreducible with 0B < ζB,and let B′ = B2/∆1,ζ . Then B′ is subdirectly irreducible and

B′/ζB′,∼= (B/ζB)2.

Proof. We write ∆ for ∆1,ζ and us the notation for congruenceson B2 introduced before Theorem 4.11. Let β be the monolith of B(smallest nonzero congruence). So we have β ≤ ζ , and ∆ + ηi = ζi,∆ · ηi = 0, (i = 0, 1) by 4.11. We claim first that (β0 ∧ β1) ∨∆ is theunique smallest congruence of B2 strictly above ∆.

To see it, first 0 6= β0 ∧ η1 ≤ β0 ∧ β1, ∆ ∧ η1 = 0, so β0 ∧ β1 6≤ ∆.Suppose that λ > ∆ in Con B2. Notice that β1 is the unique atom in1/η1 ∼= Con B, hence since 1/η1 transposes down onto η0/0, η0 ∧ β1is the unique atom below η0. Likewise for β0 ∧ η1 below η1. Now∆ ∨ (β0 ∧ η1) ≥ η0 ∧ β1 because choosing 〈x, y〉 ∈ β − 0B, we have〈x, x〉∆〈y, y〉β0 ∧ η1〈x, y〉, and hence (∆ ∨ (β0 ∧ η1)) ∧ η0 6= 0, so itcontains η0 ∧ β1. Thus if λ ∧ η1 6= 0, then λ ≥ β0 ∧ η1 and alsoλ > ∆, so λ ≥ β0 · β1. Symmetrically, we are finished with the proofof the claim unless λ ∧ η0 = λ ∧ η1 = 0. This assumption leads to acontradiction: by modularity, λ ∨ η0 = χ0 for some χ > ζ (else η0 is acommon complement of λ and ∆ in ζ0/0). [χ0, 1] = [χ, 1]0 ∧ [1, 1]1 byProposition 4.5. Also [χ0, 1] = [λ ∨ η0, η1 ∨ η0] ≤ η0 ∨ (λ ∧ η1) = η0 by

EXERCISES 57

additivity of commutator. The two equalities for [χ0, 1] yield [χ, 1] = 0,implying χ ≤ ζ , a contradiction.

Next we claim that ζ0∧ζ1 = ζ ′ is the largest γ ∈ Con B2 satisfying[γ, 1] ≤ ∆.

First, [ζ ′, 1] = 0 follows from Proposition 4.5 (in fact ζ ′ is the centerof B2). Second, suppose that γ ∈ Con B2, [γ, 1] ≤ ∆. Then [γ, 1] =[γ, η0 ∨ η1] ≤ η0 ∨ [γ, η1] and [γ, η1] ≤ [γ, 1] ∧ η1 ≤ ∆ ∧ η1 = 0. Thus[γ, 1] ≤ η0. Similarly with subscripts interchanged, so [γ, 1] = 0. By(4.40 applied to the first projection homomorphism, γ ≤ ζ0, similarlyγ ≤ ζ1. Thus the second claim is proved.

The first claim tells us that B′ is subdirectly irreducible and thesecond tells us that ζ0 ∧ ζ1 ≥ ∆ projects onto the center of B′. ThusB′/ζB′

∼= B2/(ζ0 ∧ ζ1) ∼= (B/ζ)2 obviously. �

This proposition was in the first draft manuscript of Freese-McKenzie[29],and later deleted. A much stronger result was proved by the samemethod. Let B be subdirectly irreducible with monolith β, and letγ = (0 : β) be the centralizer of β, and let σ be the centralizer of γ.Suppose that β ≤ σ < γ. Then the algebra B′ = B(γ)/∆γ,σ can beproved to be subdirectly irreducible, and we have β,≤ σ′ < γ′ wherethese congruences are defined for B′ as for B. From here, the unpub-lished argument showed that B′′ is isomorphic to a proper essentialextension of B′. This construction was iterated through the transfiniteto build a tower B(α), α an ordinal, of subdirectly irreducible alge-bras. The argument was motivated by Herrmann’s original proof ofthe fundamental theorem of Abelian algebras.

Corollaries 7.5 and 7.7 are due to C. Bergman. The fact that theinverse to the function x 7→ p(x, b, c) is a polynomial function in anilpotent variety is new.

Exercises

1. It is well known that if G is a non-Abelian group then G/Z isnot cyclic. Use Proposition 7.2 to construct a nilpotent loopL of order 9 with L/ζ ∼= Z/3.

2. Show that a nilpotent (in fact any) loop with four elements isan Abelian group.

3. Show that ifA is nilpotent and θ > 0 in Con A then θ∧ζ > 0.

4. Show that if A is a finite nilpotent algebra and B is a subal-gebra of A, then |B| divides |A|.


5. Let A be a nilpotent loop such that |A| = p2, p a prime,and |A/ζ | = p. Show that A is one-generated (but it is notnecessarily a cyclic group).

6. U, p, fn are defined before Lemma 7.3. Prove that q(x, y, z) =fn(x, z, y) is a Mal’cev term for any n–step nilpotent algebrain U.

7. Let G and H be Abelian groups and let A be their disjointunion. Define a ternary operation p(x, y, z) on A as follows. Ifx, y and z all lie in one of the groups, let p(x, y, z) = x−y+ z.Otherwise two of x, y and z lie in one group and the otherlies in the other group. In this case p(x, y, z) is the one thatis alone. Show V (A) is permutable and A is solvable. Usethis to show the results Lemma 7.3–Corollary 7.7 cannot beextended to solvable algebras.

8. Prove the first relation in Lemma 7.3.

9. Show that if A and all its homomorphic images have uniformcongruences then A has regular congruences (see Corollary7.7).

CHAPTER 8

Congruence Identities

If an equation ε of the language of lattices holds inCon A for everyA ∈ V (where V is a variety) then we write V |=con ε, and say that ε is acongruence identity of V. The most important congruence identitiesare the modular law and the distributive law (and of course this wholebook is about the consequences of congruence modularity). In theearly 1970’s it was conjectured that possibly any nontrivial lattice lawholding as a congruence identity of a variety must imply congruencemodularity. The conjecture turned out to be false (Polin [74], seealso Day-Freese [23]). In other words, there exist nonmodular varietieshaving nontrival congruence identities. As for the possible diversity ofsuch identities, and their consequences for the structure of the algebrasin a variety, our understanding is still very limited. There is an excellentsurvey article on this topic in the appendix to Gratzer [34], written byB. Jonsson. (On this topic, see also Example 11.2. and the entirety ofChapter 13.)

Recent studies of modular varieties have revealed that many pop-ular properties of a variety are intimately connected with commutatoridentities for the congruences. For example, we shall see in Chap-ter 10, Section 3 that a residually small modular variety must satisfyα ∧ [β, β] = [α ∧ β, β] (for congruences α and β in any algebra of thevariety), and that this equation is also sufficient for residual smallnessif the variety is generated by a finite algebra.

In this chapter we shall consider three special equations very muchlike the example above, as well as familiar equations like the nilpotencyand solvability laws. These equations are not written in the pure lan-guage of lattices, but rather are equations in the language of algebras〈L, x ∧ y, x ∨ y, [x, y], 0, 1〉 which are lattices with 0 and 1 and an ad-ditional binary operation [x, y]. As usual, all varieties encountered areassumed to be modular and all algebras belong to modular varieties,so these equations can be evaluated in the congruence lattices. Foran equation ε of this kind we shall write V |=con ε with just the same

59

60 8. CONGRUENCE IDENTITIES

meaning as above, and we shall consider the equation as a possible con-gruence identity of a variety. Equations involving only lattice ∨ and ∧,with no commutators, will be called pure lattice equations.

A systematic theory of congruence identities including the commu-tator does not exist at this time. Our aim here is to survey some of theinteresting results that have been discovered, and stimulate the readerto think about the subject. All of the equations we shall discuss arelisted below.

x ∧ [y, y] ≈ [x ∧ y, y](C1)

[x, y] ≈ x ∧ y ∧ [1, 1](C2)

[x, y] ≈ x ∧ y(C3)

[x, y] ≈ 0(C4)

[1, 1] ≈ 0(C5)

1, [1, 1]] ≈ 0(C6)

[[1, 1], [1, 1]] ≈ 0(C7)

[x, 1] ≈ x(C8)

For each of these equations, the property Con A |= ε is preservedunder the formation of quotient algebras and the formation of finitesubdirect products. We shall prove this fact for (C1) (Theorem 8.1) andthe reader can easily discover the proofs for the other equations. (C4) -(C7) are preserved under the formation of subalgebras and products ofany length, while each of the other equations falls to possess either oneof these preservation properties. Thus a variety generated by algebraswhose congruence lattices satisfy (Ci), (i ∈ {4, 5, 6, 7}) obeys (Ci) as acongruence identity. It can be shown, for each of the equations (C1)–(C8), that a directed union of algebras whose congruences obey theequation must itself have this property.

Equations (C4) and (C5) are equivalent for any algebra, and de-fine the class of Abelian algebras. A variety satisfying (C4) is calledAbelian. (C6) defines the two step nilpotent algebras. (See Corol-lary 7.2 for the structure of these algebras.) An equally obvious equa-tion defines the n step nilpotent algebras. For a given modular varietyV, there exists a set En of equations in the language of V such thatan algebra A is n step nilpotent if and only if A |= En. (This result

8. CONGRUENCE IDENTITIES 61

is due to Gumm; see Theorem 14.2.) (C7) defines the two step solv-able algebras. The n step solvable algebras within a given variety alsoconstitute a subvariety. We may remark that the (ordinary) equationsdefining this subvariety are much more difficult to construct than theequations for nilpotency.

Equation (C3) defines the class of congruence distributive varieties.The proof is easy (see Exercise 1). An algebra A such that Con A |=(C3) is called neutral. From the fact that finite subdirect productsand quotients of neutral algebras are neutral, it follows easily that thejoin of two distributive subvarieties of a modular variety is distributive(a result of Hagemann and Herrmann, see Exercise 2). This in turnimplies that a locally finite modular variety V has a largest distributivesubvariety. To see this, let W =

∨

i∈I Vi where Vi(i ∈ I) are thedistributive subvarieties of V. Then FW(3) is finite (a quotient of FV(3))and is a subdirect product of the FVi

(3). Hence it is a finite subdirectproduct of these algebras, and is neutral. This implies that Con FV(3)is distributive (see Exercise 1), and so V is distributive by Jonsonn’sTheorem 2.1. An example showing that modularity is a necessaryassumption for this result can be found in Chapter 11. The variety ofrings is an example of a (nonlocally finite) modular variety having nolargest distributive subvariety.

Equations (C1), (C2), and (C8) are rather different in their im-plications from the other equations. Let us consider first (C1). Thisequation implies the equation [x, [x, x]] ≈ [x, x]. Therefore. in a varietywith this congruence identity, every nilpotent algebra is Abelian, andevery nilpotent congruence is Abelian. It is known that a locally finitevariety of groups, or of rings, satisfies (C1) if and only if every nilpotentalgebra in the variety is Abelian. An algebra whose congruences obey(C1) has a largest Abelian congruences. (See Exercise 5.)

Consider an algebra A such that L = Con A satisfies (C1). Forany β ∈ L define a mapping of L into L by taking fβ(x) = [x, β]. Thenfβ(β) = [β, β] is a fixed point for fβ as we observed above. In fact,(C1) implies that fβ(x) = x for all x ≤ fβ(β). This property of thefixed points of fβ (holding for all β) is equivalent to (C1). Also eachof the equations

[x, y] ≈ (x ∧ [y, y]) ∨ (y ∧ [x, x])

[x, y] ≈ x ∧ y ∧ [x ∨ y, x ∨ y]

is equivalent to (C1). Thus if L = Con A satisfies (C1) then thecommutator in L is determined by the unary operation f(x) = [x, x]and the lattice operations.


It is not hard to show that (C1) implies (for congruence lattices ina modular variety) the equation

([1, 1] ∧ x) ∨ ([1, 1] ∧ y) ≈ [1, 1] ∧ (x ∨ y).

which with the aid of the modular law implies

([1, 1] ∨ x) ∧ ([1, 1] ∨ y) ≈ [1, 1] ∨ (x ∧ y).

The two displayed equations are actually equivalent in modular lat-tices; either of them signifies that [1,1] is a neutral element. In thesame vein. If L = Con A satisfies (C1) and β ∈ L then [β, β] is aneutral element in the interval lattice β/0A. Using these facts, A. Dayand E. Kiss [22] prove that if a locally finite nondistributive varietyV has (C1) as congruence identity, then the pure congruence identities(lattice equations) of V are the same as the pure congruence identi-ties possessed by the variety of modules over the ring of the maximalAbelian subvariety of V. (See Chapter 9 for the definition of this ring.)

We say that an algebra A satisfies a commutator equation hered-itarily if the equation holds in the congruence lattice of every subal-gebra of A. The next result was proved in the paper of R. Freese andR. McKenzie [29].

Theorem 8.1. The equation (C1) is equivalent to the implicationx ≤ [y, y] → x = [x, y]. Moreover, the class of algebras which satisfy(C1) hereditarily is closed under the formation of quotient algebras,subalgebras, and finite direct products.

Proof. First suppose that the implication holds. Clearly x ∧[y, y] ≤ [y, y]. Thus using the implication with x as x∧ [y, y] we obtainx∧ [y, y] = [x∧ [y, y], y] ≤ [x∧ y, y] ≤ x∧ [y, y]. Thus (C1) holds. Con-versely if x ≤ [y, y], then (C1) gives x = x ∧ [y, y] = [x ∧ y, y] = [x, y].

Now for the second statement. For subalgebras, the proof is trivial.For quotient algebras, let A satisfy (C1) hereditarily and C ≤ A/θ forsome θ ∈ Con A. Then C = B/γ where B ≤ A. If (C1) fails for Cthen there are µ, ν ∈ Con B, µ, ν ≥ γ, satisfying (by Remarks 4.6)ν ≤ [µ, µ]γ = [µ, µ] ∨ γ while [ν, µ]γ = [ν, µ] ∨ γ < ν. Then ν ≤ µ and[ν, µ] 6≥ [µ, µ] ∧ ν, else [ν, µ] ∨ γ ≥ ([µ, µ] ∧ ν) ∨ γ ≥ ν by modularity.Thus (C1) fails in Con B, a contradiction.

For products, suppose that C = A × B where A, B satisfy (C1)hereditarily, and D ≤ C. We can clearly assume that the projectionsmap D onto A and B, i.e. p0(D) = A, p1(D) = B . Let η1 = kerpi(i =0, 1). Now suppose that σ, µ ∈ Con D and ν = σ ∧ [µ, µ]. We shallshow that ν ≤ [ν, µ] (which implies σ ∧ [µ, µ] = [σ ∧ µ, µ]) . Sinceν ≤ [µ, µ] we have ν ∨ η0 ≤ [µ ∨ η0, µ ∨ η0]η0 . Since A satisfies (C1),


thenν ∨ η0 = [ν ∨ η0, µ ∨ η0]η0 = [ν, µ] ∨ η0.

Thus ν = ν∧([ν, µ]∨η0) = [ν, µ]∨(η0∧ν). Likewise ν = [ν, µ]∨(η1∧ν).In just the same way we can show that

(η1 ∧ ν) ∨ η0 = [(η1 ∧ ν) ∨ η0, µ ∨ η0]η0

andη1 ∧ ν ≤ [η1 ∧ ν, µ] ∨ η0.

Thus η1 ∧ ν = [η1 ∧ ν, µ] ∨ (η0 ∧ η1 ∧ ν) = [η1 ∧ ν, µ]. Therefore

ν = [ν, µ] ∨ (η1 ∧ ν) = [ν, µ] ∨ [η1 ∧ ν, µ] = [ν, µ].

�

The second statement in Theorem 8.1 holds for each of the equations(C1) - (C8). The proofs are quite similar to the one above.

We turn now to consider (C2). This equation clearly implies (C1):moreover it is rather directly implied by each of the equations (C3) and(C4). The next theorem will yield the conclusion (which is Exercise 6)that any congruence equation implied both by (C3) and (C4) is impliedalso by (C2).

Theorem 8.2. The congruence lattice of an algebra satisfies one ofthe following if and only if it satisfies all.

(1) [x, y] ≈ x ∧ y ∧ [1.1]

(2) [x, x] ≈ x ∧ [1, 1]

(3) [x, y] ≈ [x, 1] ∧ y

(4) [x ∧ y, z] ≈ x ∧ [y, z]

(5) x ≤ [1, 1]⇒ [x, x] = x

Proof. Using the commutative law of the commutator and otherbasic properties of commutators, one easily shows that (1) implies eachof (2), (3), (4) and that each of these equations implies (5). We shallnow prove that (5) implies (1). Suppose that (5) holds in Con Aand let congruences α and β be given. That [α, β] ≤ α ∧ β ∧ [1, 1] isclear. Let σ = α ∧ β ∧ [1, 1]. Thus σ = [σ, σ] by (5), and it’s clear that[σ, σ] ≤ [α, β]. Thus we conclude that (1) holds. �

Theorem 8.3. A variety V has (C2) as congruence identity if andonly if every non-Abelian subdirectly irreducible algebra in V has a non-Abelian monolith.


Proof. In one direction (necessity of the condition) this is imme-diate from the last theorem (the equivalence of (1) and (5)). For theconverse (sufficiency of the condition), assume that V does not satisfy(C2). Then we can choose A ∈ V and a congruence λ of A such that[λ, λ] < λ ≤ [1, 1] (by the previous theorem). There exists a congru-ence σ such that [λ, λ] ≤ σ, λ 6≤ σ, and A/σ is subdirectly irreducible.Let σ > σ be the congruence corresponding to the monolith of A/σ.Then λ ∨ σ ≥ σ, so that

[σ, σ] ≤ [λ, λ] ∨ σ ≤ σ,

implying that the monolith of A/σ is Abelian. (See Proposition 4.4.)Since [1, 1] 6≤ σ, we also have that A/σ is non-Abelian. �

A congruence β of an algebra is called prime if and only if β ≥ [σ, γ]always implies β ≥ σ or β ≥ γ. An algebra is called prime if andonly if the commutator of any two of its non-zero congruences is non-zero. Theorem 8.3 says that a variety satisfies (C2) if and only if itssubdirectly irreducible algebras are Abelian or prime. It is known thata locally finite variety of groups having this property must be Abelian.However, quite a few well studied properties of varieties are known toimply (C2). The equation arose first in the investigation of locally finitevarieties with decidable first order theory (in Burris-McKenzie [8]). Allsuch decidable varieties satisfy (C2). Semi-simple varieties (in whichthe subdirectly irreducible algebras are simple) obviously satisfy (C2).(It follows from Theorem 8.3).

An algebra A is said to have the congruence extension property ifand only if for every subalgebra B ≤ A and congruence β on B, there isa congruence α on A for which α|B = β. A variety has the congruenceextension property if all of its algebras have the property. E.Kiss [55]has proved that every modular variety with the congruence extensionproperty satisfies (C2). Our next theorem (and Theorem 8.2) are dueto him.

Theorem 8.4. Suppose that A ×A has the congruence extensionproperty. Then Con A |= (C2). Moreover, if α, β ∈ Con A andB ≤ A the [α|B, β|B] = [α, β]|B.

Proof. We assume throughout the argument that A2 has the con-gruence extension property. To prove that (C2) holds, we use the equiv-alent from (3) from Theorem 8.2. Let α, β ∈ Con A. We are to showthat [α, β] ⊇ [α, 1]∧ β. Let ∆β,α be the congruence on A(β), and ∆1,α

be the congruence on A2, defined in Definition 4.7. These congruencesare generated in the respective algebras by the same set. Thus since A2

has the congruence extension property, it follows that ∆1,α|A(β) = ∆β,α.


Now suppose that 〈x, y〉 ∈ [α, 1]∧ β. We can employ Theorem 4.9. Wehave that 〈x, y〉, 〈x, x〉 ∈ A(β) and 〈x, y〉 ≡ 〈x, x〉 (mod ∆1,α). There-fore 〈x, y〉 ≡ 〈x, x〉 (mod ∆β,α), implying that 〈x, y〉 ∈ [α, β] as desired.

To prove the other commutator property (restriction of commuta-tors), let α and β be as before, and let B ≤ A. We first observe that ourhypothesis implies that B2 has the congruence extension property, andthus by what was proved, the congruences inB satisfy (C2). Using (C2)in both algebras, we see that it will suffice to prove the restriction prop-erty in the case α = 1A = β : i.e. to prove that [1A, 1A]|B = [1B, 1B].Let λ be the congruence on A generated by the set 1B. Clearly

∆1A,λ = CgA2(∆1B ,1B ).

Thus

∆1B ,1B = ∆1A,λ|B2

implying that

[1A, λ]|B = [1B, 1B].

Now using (C2) for A and the fact that λ ≥ 1B we get the desiredequation [1A, 1A]|B = [1B, 1B]. �

Several of the useful properties possessed by the commutator inevery modular variety can be written as congruence identities, namely

[x, y] ≈ [y, x] ≈ x ∧ y ∧ [x, y].

and

[x, y ∨ z] ≈ [x, y] ∨ [x, z].

Although (C3) is a congruence identity of every distributive variety,it is easy to show that it cannot be derived from the basic equationsabove and the equations defining distributive lattices in the same waywe proved Theorem 8.2 (using only the machinery of equational logic).Likewise, (C8) is equivalent for varieties to the equation [1, 1] ≈ 1; andthis equivalence is a theorem not of equational logic, but of commutatortheory.

Theorem 8.5. The following are equivalent for a variety V.

1. V |=con [x, 1] ≈ x.2. V |=con [1, 1] ≈ 1.3. V does not possess nontrivial Abelian algebras.4. Every algebra A ∈ V is centerless, i.e., ζA = 0A.5. If A, B ∈ V, then Con(A×B) ∼= Con A×Con B naturally.

Proof. The equivalences (1) ⇐⇒ (4) and (2) ⇐⇒ (3) are imme-diate. Moreover (4) ⇒ (3) is trivial. The (surprising) implication (3)


⇒ (4) is by Proposition 7.1. Thus (1)–(4) are seen to be mutuallyequivalent.

Next we show that (5) ⇒ (1). Suppose that we have A ∈ V andβ ∈ Con A and (5) holds. The meaning of (5) is that every congruenceof A2 should be of the form γ0∧δ1, where γ and δ are congruences of Aand 〈x, y〉 ≡ 〈u, v〉 (mod γ0) if and only if 〈x, u〉 ∈ γ, and 〈x, y〉 ≡ 〈u, v〉(mod δ1) if and only if 〈y, v〉 ∈ δ. Now suppose that ∆1A,β = γ0 ∧ δ1.Since 〈x, x〉 ∆1A,β 〈y, y〉 whenever 〈x, y〉 ∈ β it follows that β ≤ γ ∧ δ.But now β0∧β1 ≤ γ0∧δ1 = δ1A,β. This certainly implies that [β, 1A] =[1A, β] = β by Theorem 4.9.

We conclude the proof by showing that (1) ⇒ (5). Suppose that(1) holds and that θ ∈ Con(A×B) with A, B ∈ V. We write η0, η1 forthe kernels of the projection homomorphisms. It will suffice to showthat θ = θ where θ = (η0 ∨ θ) ∧ (η1 ∨ θ). Now

[1, θ] ≤ [η0 ∨ η1, η0 ∨ θ] ≤ η0 ∨ (η1 ∧ θ).

and similarly [1, θ] ≤ η1 ∨ (η0 ∧ θ). Thus

θ = [1, θ] ≤ (η0 ∨ (η1 ∧ θ)) ∧ (η1 ∨ (η0 ∧ θ)) ≤ θ

by two applications of the modular law and the fact that η0∧η1 ≤ θ. �

Remark 8.6. It is an immediate consequence of the theorem thata modular variety obeys (C8) if and only if the finite products of itsalgebras possess no skew (see Exercise 3) congruences. Compare thiswith the fact that a modular variety obeys (C3) (is distributive) if andonly if the finite subdirect products of its algebras possess no skewcongruences. The variety of rings with unit obeys (C8). See Exercise 8for an interesting application of (C8): An algebra whose congruencesobey (C8) has the refinement property (Chang-Jonsson-Tarski [11]) fordirect decompositions, and so can be represented in at most one wayas a direct product of directly indecomposable algebras.

Remark 8.7. A variety is distributive if and only if all of its alge-bras generated by three elements have distributive congruence lattices.A variety is modular if and only if all of its algebras generated by fourelements have modular congruence lattices. These facts are immediateconsequences of Theorems 2.1 and 2.2. Much more generally, it canbe proved that every congruence equation expressed in the operationsx∨ y, x∧ y, [x, y] and 0 and 1 is local - holds in a modular variety V ifand only if it holds in every finitely generated algebra in V.

Remark 8.8. We remarked that each of the equations (C1)–(C8) ispreserved under the formation of quotient algebras. This is not true of


every congruence equation, as may be seen by considering this equationwhich is weaker than (C1):

[x, x] ∧ [y, y] ≤ [x, y].

Let G be a group with |Z(G)| = 2, G/Z(G) non-Abelian, and Z(G)the only nontrival normal subgroup ofG, for exampleG = SLProposition 4.3.This information determines the commutator operation on Con G.Now Con(G×G) is the lattice of Figure 1.

0

1

π

η1

α1

η0

α0

Figure 1.

Every congruence of G × G is a direct product congruence ex-cept π. Thus the commutator on Con(G × G) is determined byProposition 4.5(3), and by the fact the π is in the center. For ex-ample, [α0, α0] = η0. Using these facts, the reader can show that theequation above holds in Con(G×G). However, it fails in (G×G)/π(take x = α0/π and y = α1/π).

Remark 8.9. Each of the equations (C1)–(C8) is preserved underfinite subdirect products, and also had the property that it holds asa congruence identity of a variety V if and only if it holds in the con-gruence lattice of each subdirectly irreducible algebra in V. Now if εis a congruence equation that is preserved by taking quotient algebrasand finite subdirect products then, since the validity of ε is in addi-tion a local property, we can conclude that for any finite algebra F,V (F) |=con ε if and only if F satisfies ε hereditarily. In particular, itfollows that a finite algebra F generates a distributive variety if andonly if F is hereditarily neutral (satisfies (C3) hereditarily). This is aresult of Hagemann-Hermann [44].

Problem 8.10. Find some simple commutator equations that holdin every modular variety, but do not follow in equational logic from the


equations defining modular lattices and the equations expressing com-mutativity and additivity of the commutator, and the equation [x, y] ≤x∧ y. (The Arguesian equation of B. Jonsson is one example, but it isa pure equation, without commutator, and is rather complex.)

Problem 8.11. Is it true that for every modular and nondistribu-tive variety V, the pure congruence identities are the same as the purecongruence identities possessed by the variety of modules over the ringof the maximal Abelian subvariety of V?

This problem was refuted by P. P. Palfy and C. Szabo in [69] and[68]; see Chapter 15 where some positive results along these lines arealso presented.

Exercises

1. Show that if Con A |= [x, y] ≈ x ∧ y then Con A is adistributive lattice.

2. An algebra whose congruence lattice obeys (C3) is said to beneutral. Prove that a subdirect product of finitely many neu-tral algebras is neutral. Consequently, the join of two distribu-tive subvarieties of a modular variety is distributive.

3. If A ⊆ A0 ×A1 is a subdirect product, a congruence α of Ais said to be skew if and only if α is not of the form γ0 ∧∆1

(with γ ∈ Con A0 and ∆ ∈ Con A1). Prove the equivalenceof the following statements for any modular variety. (See theproof of Theorem 8.5.)(i) V is distributive,(ii) V |=con (C3) ,(iii) Subdirect products of two (or of n) algebras in V have no

skew congruences.

4. Prove that in any modular lattice L with 0 and 1, an elementa satisfies a∧ (x∨ y) = (a∧ x)∨ (a∧ y) for all elements x andy if and only if a satisfies (a∨ x) ∧ (a∨ y) = a∨ (x∧ y) for allelements x and y if and only if the mapping x→ 〈a∨x, a∧x〉is a subdirect embedding of L into the product of its intervalsublattices 1/a and a/0. (Such an element a is called neutral.)Conclude that if Con A |= (C1) then Con A is subdirectlyembeddable into (1A/[1A, 1A])× ([1A, 1A]/0A).

EXERCISES 69

5. Show that if Con A |= (C1) the the join of two Abeliancongruences of A is Abelian. Then conclude that A has alargest Abelian congruence.

6. Show that a congruence equation is implied by (C2) if andonly if it is implied by each of (C3) and (C4). [There is a wayto use the result of Exercise 4. This exercise is true whether“imply” is construed in the sense of equational logic (the basiccommutator equations and the modular law being implicatlyassumed) or in the sense that every variety obeying the one asa congruence equation obeys also the other.]

7. Show that rings with unit obey (C8).

8. Suppose that Con A |= (C8). Let α and β be complementsin Con A such that α ◦ β = β ◦ α. Show that the mappingλ→ β∨λ is an endomorphism of Con A that commutes withthe complete lattice operations. (This is essentially the strictrefinement property of Chang-Jonsson-Tarski.)

9. Let k be a positive integer not less than 3. Let 〈V,+,−, 0〉 bea group of cardinality 2k satisfying x + x = 0 (i.e. a vectorspace of dimension k over the 2 element field). Let · and ⊕be any operations such that 〈V, ·,⊕〉 is a lattice. Let A be thealgebra whose universe is V and basic operations are +,−, fdefined by

f(x0, . . . , xk+1) =

xk · xk+1 if x0, . . . , xk−1 are

linearly independent

0 otherwise

and a k + 2-ary operation g defined the same way from ⊕.Prove that A is neutral, but that every algebra generated byk − 1 or fewer elements in V (A) is Abelian.

10. Prove that a variety of groups has the congruence extensionproperty if and only if the variety consists of Abelian groups.

CHAPTER 9

Rings Associated With Modular Varieties:

Abelian Varieties

Let V be a modular variety and let A be all Abelian algebras in V.By Proposition 4.4 and Proposition 4.5, A is a subvariety of V. LetA ∈ V, and let β ∈ Con A be an Abelian congruence. Recall that,by Corollary 5.8, each block of β is an Abelian group with additionx + y = d(x, z, y), where z is any fixed element of the block and ofcourse d is a Gumm difference term. Moreover the unary polynomialsof A induce affine maps (i.e., maps which preserve d(x, y, z)) betweenthe β blocks. In this chapter we will construct matrix rings for V, actingon the direct sum of the blocks, which capture these affine actions. Inthis way the direct sum of the blocks becomes a module over one ofour rings. There is a strong connection between the structure of thismodule and the structure ofA (see, for example, Theorem 9.9). Severalof the results of this chapter will be applied in the subsequent chapters.

In the case V is Abelian, i.e. V = A, only one of these rings is neededand the variety V is closely related (in fact polynomially equivalent) tothe variety of modules over this ring. The connection, which is actuallystronger than this, is established at the end of the chapter.

Before commencing with the constructions and proofs, let us re-mark that affine algebras and varieties composed of then have beenstudied, for different reasons, by a number of authors since Ostermanand Schmidt first (it seems) gave them serious consideration in [67].The papers of Bela Csakany, [16] and [17], and Gumm’s [37], havebetween them a good set of references to this literature. Chapters 4and 5 of Smith’s book [79] should also be consulted. In our opinion thetreatment presented here is essentially novel, and more satisfactory inseveral respects than the previous treatments. (Our readers, of course,will have their own opinions.) Nevertheless, the subject is not as easy asit first looks and work remains to be done. As stated above the resultsof this chapter will be useful in some of the applications presented inthe later chapters. The detailed description of Abelian varieties givenat the end of the chapter has been applied in Baldwin-McKenzie [1],Burris-McKenzie [8], and McKenzie [63].

71

72 9. RINGS ASSOCIATED WITH MODULAR VARIETIES

Let A ∈ V, z, z′ ∈ A and let β be an Abelian congruence on A.Recall that M(β, z) is the β block, z/β, of A containing z and that itis an Abelian group under x+y = d(x, z, y) with z as the zero element.Let Hom(β, z, z′) denote the set of functions g : M(β, z)→M(β, z′) ofthe form

(1) g(x) = f(x, z, z′, c0, . . . , ck−1)

where c0, . . . , ck−1 ∈ A and f is a k + 3-ary term, k < ω, such that Vsatisfies

(2) f(v, v, v′, y0, . . . , yk−1) ≈ v′.

The next lemma gives a much simpler description of Hom(β, z, z′)and shows that it does not depend on V.

Lemma 9.1. Hom(β, z, z′) is the set of restrictions to z/β of unarypolynomials on A which map z to z′.

Proof. Let h be a (k + 1)-ary term such that

h(z, c0, . . . , ck−1) = z′.

Define a term f by

f(u, v, v′,y) = d(h(u,y), h(v,y), v′)

where y = (y0, . . . , yk−1). Clearly V satisfies (2) and, since β is Abelian.

f(x, z, z′, c0, . . . , ck−1) = h(x, c0, . . . , ck−1)

for all x ∈M(β, z) by Theorem 5.5(ii) �

By Corollary 5.8 each g ∈ Hom(β, z, z′) is a group homomorphismfrom M(β, z) into M(β, z′). Hom(β, z, z′) is an Abelian group underthe operation (g + h)(x) = d(g(x), z′, h(x)), for g, h ∈ Hom(β, z, z′). Ifz′′ is another element ofA and if g ∈ Hom(β, z, z′) and h ∈ Hom(β, z′, z′′)then we define hg by hg(x) = h(g(x)), of course. As an exercisethe reader can show that the distributive laws hold. In particular,Hom(β, z, z) is a ring with 1. Also note that if f(x1, . . . , xn) is a polyno-mial on A and f(z1, . . . , zn) = z then f defines a group homomorphismof M(β, z1)×· · ·×M(β, zn) into M(β, z), by Corollary 5.8. Moreover,if we let gi(xi) = f(z1, . . . , zi−1, xi, zi+1, . . . , zn) then gi ∈ Hom(β, zi, z)and

(3) f(x1, . . . , xn) =n

∑

i=1

gi(xi).

If R is the ring of n by n matrices whose (i, j)th elements lie inHom(β, zj, zi), then this ring acts naturally onM(β, z1)×· · ·×M(β, zn)

9. RINGS ASSOCIATED WITH MODULAR VARIETIES 73

and by the equation displayed above a great deal of information aboutthe algebraic structure of A is contained in this action.

In this chapter we construct a class of matrix rings for V which willact on the direct sum of some of the blocks of an Abelian congruence.The class will have two parameters: λ, the size of the matrices, and κ,the number of constants. Thus let X = {u}∪{vi : i < λ}∪{yi : i < κ},where λ ≥ 1 and κ ≥ 0 are cardinals and let F = FV(X). Let eidenote the endomorphism of F which sends u to vi and fixes all othergenerators. Let θi = Cg(u, vi).

Definition 9.2. Let Hij = {r ∈ F : eir = vj} and let Hij ={r/[θi, θi] : r ∈ Hij}.

If r = r(u,v,y) and s = s(u,v,y) are in Hij and t = t(u,v,y) isin Hjk then we define

r + s = d(r, vj, s)

−r = d(vj, r, vj)

t ◦ r = t(r,v,y)

(4)

Let πr denote the canonical projection of F onto F/[θi, θi]. Operationson the Hij’s are defined by

πi(r) + πi(s) = πi(r + s)

−πi(r) = πi(−r)

πj(t) ◦ πi(r) = πi(t ◦ r)

(5)

Notice that Hij is simply the set of those elements of F which canbe represented by a term r(u, v, y) such that r(vi, v, y) ≈ vj holds in V.It is easy to see that the above operations on the H ′

ijs are well definedexcept possibly for composition where the problem is to show that ift [θj , θj] t

′, for some t, t′ ∈ Hjk, and r ∈ Hij then t ◦ r [θi, θi] t′ ◦ r.

Let ℓ be the endomorphism of F which sends u to r and fixes all othergenerators. Note that eiℓ = ej (since r ∈ Hij) and that ker ek = θk.Thus ℓ−1(θi) = θj . Indeed, (a, b) ∈ ker ej if and only if eja = ejb ifand only if eiℓa = eiℓb if only if (ℓa, ℓb) ∈ ker ei if and only if (a, b) ∈ℓ−1(ker ei). Now using Proposition 4.4(2) and then Proposition 4.4(1)we have

ℓ−1[θi, θi] ≥ ℓ−1[θi|rng ℓ, θi|rng ℓ]

= ℓ−1[ℓℓ−1(θi), ℓℓ−1(θi)]

= [ℓ−1(θi), ℓ−1(θi)] ∨ ker ℓ

≥ [θj , θj]


Hence ℓ[θj , θj ] ⊆ [θi, θi]. Now t ◦ r = ℓ(t) and t′ ◦ r = ℓ(t′). Thust ◦ r [θi, θi] t

′ ◦ r, as desired.It follows easily from Proposition 5.7 that Hij is an Abelian group

with πi(vj) as null element. The composition satisfies the associativeand distributive laws. For example, if r, s ∈ Hij and t ∈ Hjk then sincer θi vj θi s, we have by Proposition 5.7

t ◦ (r + s) = t(r + s,v,y)

= t(d(r, vj , s),v,y)

≡ d(t(r,v,y), t(vj,v,y), t(s,v,y)) (mod [θi, θi]))

= d(t ◦ r, vk, t ◦ s)

= t ◦ r + t ◦ s,

proving the left distributive law.In particular Hii is a ring with 0 = πi(vi) and 1 = πi(u). Also, by

Proposition 5.7, one can show that the group Hij does not depend onthe choice of d(x, y, z), so long as it satisfies Theorem 5.5(i,ii).

Definition 9.3. Let R(V, λ, κ) be the ring of all λ by λ matrices(mij) withmij ∈ Hji such that each column contains only finitely manynonzero entries. Addition and multiplication are the ordinary matrixoperations. The zero of this ring is the matrix whose (i, j)th element isπjvi. The unit element of this ring has (i, j)th element πjvi, for i 6= j,and πiu for i = j. We let R(V, λ) = R(V, λ, 0) and R(V) = R(V, 1).

Now let A ∈ V, β an Abelian congruence on A and zi ∈ A, i < λ.Let M =

∑

i<λM(β, zi) denote the direct sum of the Abelian groupsM(β, zi), i.e., M = {(ai)i<λ : ai = zi for all but finitely many i’s}.For each choice of constants ck ∈ A, k < κ, we make M into anR(V, λ, κ)–module. Let (mij) ∈ R(V, λ, κ) and let mij = πjrij ∈ Hji,with rij = r(u,v,y) ∈ Hji. If aj ∈ zj/β then we define

(6) mij · aj = (πjrij) · aj = r(aj, z, c).

More formally, this means (πjrij) · aj is the image of rij under the ho-momorphism σ : F → A sending u to aj, vk to zk, and yk to ck. Sincerij ∈ Hji, r(vj ,v,v) ≈ vi holds in V, and hence the map aj → (πjrij)·aj,defined above, is an element of Hom(β, zjzi) by Lemma 9.1. More-over, looking at the proof of Lemma 9.1, we see that every element ofHom(β, zj, zi) can be represented in this way (provided κ is sufficientlylarge).

In order to show that the multiplication given by (6) is well defined,we need to check that the map given above is well defined. That is,we need to show that it is independent of the choice of rij . To do this


we must show that if r, r′ ∈ Hij and r(u,v,y) [θj , θj ] r′(u,v,y) then

r(aj, z, c) = r′(aj, z, c) for aj ∈ zj/β. To see this let σ : F → A bethe homomorphism defined above and let π be the kernel of σ. Nowσ(θj + π) = σ(Cg(u, vj) + π) = Cg(σ(u), σ(vj)) = Cg(aj , zj) ≤ β. ByProposition 4.4(1)

σ([θj , θj] + π) = [σ(θj + π), σ(θj + π)]

≤ [β, β]

= 0

Thus, since r(aj, z, c) and r′(aj , z, c) are related by the congruence on

the left, they are equal.Now if m = (mij) is a matrix in R = R(V, λ, κ) and a = (ai) ∈

∑

M(β, zi) we define m ·a = b, where b = (bi) with bi =∑

j<λmij ·aj,that is, m acts by ordinary matrix multiplication, where we view aand b as column vectors. Since mij · aj ∈ M(β, zi) and, for all butfinitely many j, aj = zj, the sum defining bi makes sense. Moreover,since m is column finite, bi = zi for all but finitely many i, so thatb ∈

∑

M(β, zi). To see that M is a module under this action we needto check four things. First it is easy to see that if 1 is the identityof R(V, λ, κ) then 1 · a = a. If n = (nij) is also in R, we need toshow that (m · n) · a = m · (n · a). This follows from the fact that(mik · nkj) · aj = mik · (nkj · aj), which in turn follows easily from thedefinitions. (Essentially this is just the fact that function compositionis associative.) We also need to show that both distributive laws hold.The law (m+ n) · a = m · a+ n · a follows directly from the definitionand the argument for the other distributive law is similar to the proofof the distributive law given above. The details are left to the reader.

Notice that, in particular, M(β, z) is an R(V, 1, κ)–module for eachchoice of constants ck ∈ A, k < κ.

Arbitrary polynomials as well as unary polynomials are closely re-lated to the sets Hom(β, zi, zj). Indeed, if f(x1, . . . , xn−1) is a polyno-mial on A with f(z1, . . . , zn−1) = z0 then

gi(x) = f(z1, . . . , zi−1, z, zi+1, . . . , zn−1) ∈ Hom(β, zi, z0).

Hence there is an mi ∈ Hi0 with gi(x) = mi · x. Thus we have therepresentation

(7) f(a1, . . . , an−1) =

n−1∑

i=1

mi · ai, ai ∈ zi/β.


As an example, let λ = 1 and let V be the variety of groups.An Abelian congruence of G ∈ V corresponds to an Abelian nor-mal subgroup H. We will use v in place of v0 since there is onlyone v. For k < κ, let rk(u, v,y) = ykuv

−1y−1k v. Since rk(v, v,y) ≈ v,

rk(u, v,y) ∈ H00. If gk ∈ G are fixed and we choose the correspondenceyk → gk and let z (the image of v in G) be the identity of G, and ifx ∈ H, then (π0rk) · x = g−1

k xgk. Thus the action on H of conjugationby a set of elements ofG is included in theR(V, 1, κ) action onH. Thisaction is extremely important in group theory. The ring R(V, 1, 0) doesnot contain this action on H.

Before we begin our study of Abelian congruences, we prove thereeasy results about R(V, λ, κ).

Proposition 9.4. If κ ≤ κ′ then R(V, λ, κ) is both a subring anda homomorphic image of R(V, λ, κ′).

Proof. Let X be as before and let X ′ = {u} ∪ {vi : i < λ} ∪ {yk :k < κ′}, and let F′ = FV(X

′) and view F ⊆ F′. Let ϕ : F′ → F mapyk to v0 for κ ≤ k < κ′ and fix all other generators. Clearly Hij ⊆ H ′

ij

and ϕ(H ′ij) = Hij. Now let θ′i = CgF′(u, vi) and note θi = θ′i ∩ F2 and

ϕ(θ′i ∨ kerϕ) = θi. By Proposition 4.4(1) ϕ([θ′i, θ′i] ∨ kerϕ) = [θi, θi]

and from this and Proposition 4.4(2) we obtain [θi, θi] = [θ′i, θ′i] ∩ F2.

From this it follows that for r, s ∈ F, r [θi, θi] s if and only if r [θ′i, θ′i]

s and hence the natural map from Hi into Hij is well defined andinjective. Since ϕ(θ′i) = θi, ϕ([θ

′i, θ

′i]) ⊆ [θi, θi]. From this it follows

that there is a natural map ϕ from Hij onto Hij . These injections canbe combined into an injection of R(V, λ, κ) into R(V, λ, κ′) which is aring homomorphism. Similarly the natural surjections can be combinedinto a ring surjection of R(V, λ, κ′) onto R(V, λ, κ). The details are leftto the reader. �

Theorem 9.5. Let V be a modular variety and A its maximalAbelian subvariety. Then the following are equivalent

(i) V is distributive,(ii) |R(V, λ, κ)| = 1 for all λ ≥ 1 and κ ≥ 0,(iii) |R(V, 1, 0)| = 1,(iv) |R(V, λ, κ)| = 1 for some λ ≥ 1 and κ ≥ 0.

Moreover, V satisfies (C8) (of Chapter 8) if and only if |R(A, 1, 0)| = 1.

Proof. If V is distributive then [θi, θi] = θi. Hence Hij = {πi(vj)}and so |R(V, λ, κ)| = 1. If (iv) holds and β is an Abelian congruenceon A ∈ V, then

∑

i<λM(β, zi) is an R(V, λ, κ)–module. This forcesM(β, z) to be trivial for z ∈ A. Hence β = 0. This easily implies thatV is distributive.


Clearly (ii) implies (iii) implies (iv). The last statement followsfrom the fact that V satisfies (C8) if and only if A is trivial. But anAbelian variety is trivial if and only if it is distributive. �

Proposition 9.6. If U is a subvariety of V then R(U, λ, κ) is ahomomorphic image of R(V, λ, κ).

Proof. If U is the trivial variety that result is clear. So assumethat U is nontrivial and let f be the homomorphism of FV(X) ontoFV(X) which is the identity on X . This clearly induces a map fromHij(V) to Hij(U). The proof of Lemma 9.1 shows this map is onto. ByProposition 4.4(1) this induces a map of Hij(V) onto Hij(U). It is easyto see that this map respects addition and composition from which itfollows that R(U, λ, κ) is a homomorphic image of R(V, λ, κ). �

We now look more closely at the ring action on an Abelian congru-ence β of an algebra A. Let zi, i < λ, and ck, k < κ, be elements of A.Recall that if r(u,v,y) ∈ Hij Then the restriction of g(x) = r(x, z, c)to zi/β is in Hom(β, zi, zj). When can every element of Hom(β, zizj)be represented in this way? It follows from Lemma 9.1 that his will bethe case if there are enough constants. One way to ensure this is tochoose a constant for each element of A (so that κ = |A|). Althoughthe example of the Abelian normal subgroup before Pproposition 9.4shows that we do need the constants for our ring, we will see that inmany important cases we can get by without them. We will see belowin Theorem 9.12 that they are not necessary at all for Abelian varieties.

Let β an Abelian congruence on A and let zi ∈ A, i < λ. Chooseconstants ck, k < κ, and another set of constants c′k, k < κ. Thesesets of constants determine two ring actions on

∑

M(β, zi). The nextresult relates these actions.

Theorem 9.7. Let β, zi, i < λ, ck, c′k, k < κ, be as described

above. Let γ = (0 : β) be the centralizer of β. If ck γ c′k, for k < κ,

then the two ring actions of R(V, λ, κ) on∑

M(β, zi) determined bythe two sets of constants are the same.

Proof. Let r(u,v,y) be a term representing an element of Hij , sothat r(vi,v,y) ≈ vj is an identity of V. Then in A r(zi, z, c) = zj =r(zi, z, c

′) . Since c γ c′ and [β, γ] = 0, the term condition implies thatr(ai, z, c) = r(ai, z, c

′) for ai ∈ zi/β. From this it follows easily thatthe two ring actions are the same. �

Again suppose β is an Abelian congruence on A and let γ = (0 : β)be its centralizer. Let zi ∈ A, i < λ and suppose that each γ–class isrepresented by some zi. (Notice that this is weaker than the assumption


that each β–class is represented.) In this case we will not need anyck’s; the zi’s will serve as constants. To see this suppose that ck, k < κ,are elements of A. Let x = {u} ∪ {vi : i < λ} ∪ {yk : k < κ}and X0 = {u} ∪ {vi : i < λ}. Then M =

∑

M(β, zi) is both anR = R(V, λ, κ) and an R0 = R(V, λ) = R(V, λ, 0) module. By thedefinition of a module this means that there are ring homomorphismsfrom bothR andR0 into the ring of endomorphisms ofM as an Abeliangroup. Using the last theorem we can see that the images of these twohomomorphisms are the same and thus the two modules RM and R0

Mare the same. The idea is this. If r(u,v,y) is a term representingand element in Hij then let s(u,v) be the term obtained from r byreplacing each yk with vm, where m is the (first say) index such thatck γ zm. As in the last theorem it is easy to see that s(u,v) ∈ H0

ij

(the Hij associated with X0) and that the elements of Hom(β, zi, zj)determined by r and s are the same, i.e., (πir)·x = (πis)·x for x ∈ zi/β.Thus in summary, if β is an Abelian congruence on A with centralizerr = (0 : β) and zi ∈ A, i < λ, represents all the γ–classes, thenthe actions of R(V, λ, κ) and of R(V, λ) on

∑

M(β, zi) are the same.Thus in studying

∑

M(β, zi) we take R(V, λ, 0) as our ring. For futurereference we record the fact that each element of Hom(β, zi, zj) can berepresented by an element of H0

ij .

Lemma 9.8. With A, β, γ, zi, i < λ and H0ij as above, let h ∈

Hom(β, zi, zj). Then there is an element r ∈ H0ij (and so representable

by a term r(u, v), with r(vi, v) ≈ vj in V) such that h(x) = (πir) · x =r(x, z) for all x ∈ zi/β. �

The map described above, which sends the yk to vi where i is thefirst index such that ck γ zi, induces a ring homomorphism of R ontoR0. In Exercise 6 you are asked to show that the kernel of this ringhomomorphism is contained in the kernel of the action of R on M.

The next theorem shows that there is a strong connection betweenA and the module

∑

M(β, zi).

Theorem 9.9. Let A ∈ V and let β ∈ Con A be an Abeliancongruence. Let γ ∈ Con A satisfy γ ≥ β and [β, γ] = 0, i.e., β ≤γ ≤ (0 : β). Let zi, i < λ, be elements of A such that each γ–classcontains at least one zi. Then the lattice of submodules of the module∑

M(β, zi) over R(V, λ) is isomorphic to the interval β/0 in Con A.

Proof. We now let R = R(V, λ) (= R(V, λ, 0)) and take H0ij to

be as in the previous lemma. If δ ≤ β let M(δ) =∑

i<λM(δ, zi). Itis easy to see that M(δ) is a submodule of M = M(β), and the mapδ 7→M(δ) is order-preserving.


For the inverse map suppose that N is a submodule of M. Definea relation αN on A by x αN y if and only if two conditions hold.First xβy. Since β ≤ γ this implies that xγy and thus there is ani such that x, y,∈ zi/γ. For j < λ, let ιj : M(β, zj) → M be themap which sends aj ∈ M(β, zj) to the vector whose jth coordinate isaj and whose other coordinates are 0 (i.e., whose kth coordinate is zk,for k 6= j). The second condition is that ιid(x, y, zi) is in N. This lastcondition is equivalent to saying that there is some element of N whoseith entry is d(x, y, zi). Moreover, there could be more than one i withx, y ∈ zi/γ, but the definition is independent of which i is used. Tosee the former statement let a ∈ N have ith entry d(x, y, zi). Let mbe the matrix in R whose (i, i)th entry is 1(= πi(u)) and whose otherentries are 0. Then ιid(x, y, zi) = m · a ∈ N, proving the first claim.To see the second statement suppose that x, y ∈ zi/γ and x, y ∈ zj/γ.Let r = d(u, vi, vj) ∈ Hij . It follows easily form Proposition 5.7 that(πir)·d(x, y, zi) = r(d(x, y, zi), zi, zj) = d(x, y, zj). Let m be the matrixin R whose (j, i)th entry is πi(r) and whose other entries are 0. Ifa = ιid(x, y, zi) and b = ιjd(x, y, zj) then m · a = b, which proves thesecond claim.

Using Proposition 5.7 and the Abelian group operations on N , onecan show that αN is an equivalence relation on A. To see that it isa congruence let g be a unary polynomial on A. Assume that x αN yand that x, y ∈ zi/γ. We need to show that g(x) αN g(y). Choose jso that g(zi) ∈ zj/γ. Let h(w) = d(g(w), g(zi), zj). Then h(zi) = zj ,so h ∈ Hom(β, zi, zj) by Lemma 9.1. By the previous lemma, thereis an element r ∈ H0

ij such that (πir) · c = h(c) for all c ∈ zi/β. Letx αN y and let a = ιid(x, y, zi). Then a ∈ N by the definition ofαN. Let m ∈ R be the matrix whose (j, i)th entry is πir and whoseother entries are all 0. Since N is a submodule, m · a ∈ N. Allcomponents of m · a are 0 except the jth which is h(d(x, y, zi)). Nowsince x β y γ zi, Proposition 5.7 yields h(d(x, y, zi)) = d(h(x), h(y), zj).Thus ιjd(h(x), h(y), zj) ∈ N. This shows that h(x) αN h(y). Now wecalculate

d(h(x), h(y), zj) = d(d(g(x), g(zi), zj), d(g(y), g(zi), zj), d(zj, zj , zj))

= d(d(g(x), g(y), zj), d(g(zi), g(zi), zj), d(zj, zj , zj))

= d(d(g(x), g(y), zj), zj, zj)

= d(g(x), g(y), zj)


The second equality follows from Proposition 5.7 ”backwards” sinceg(x) β g(y) γ zj and g(zi) γ zj . The last equality is valid since

d(g(x), g(y), zj) β d(g(x), g(x), zj) = zj .

Thus g(x) αN g(y). Hence αN is a congruence. Clearly, αN ≤ β andthe map N→ αN preserves order.

We now have two order-preserving maps between the lattices. Weneed to show that they are mutually inverse, i.e., we need to show thatfor δ ≤ β, δ = αM(δ), and for N a submodule of M, N = M(αN). Tosee the first, let xβy and x, y ∈ zi/γ. Then since xβy,

x αM(δ) y ⇔ ιid(x, y, zi) ∈M(δ)

⇔ d(x, y, zi) δ zi

⇔ xδy.

That d(x, y, zi) δ zi implies x δ y follows from the fact that x =d(d(x, y, zi), zi, y) which follows Proposition 5.7 since

d(d(x, y, zi), zi, y) = d(d(x, y, zi), d(y, y, zi), d(y, y, y))

= d(d(x, y, y), d(y, y, y), d(zi, zi, y))

= d(x, y, y) = x.

Now if a ∈ M and N is a submodule, then a ∈ N if and only ifιiai ∈ N for each i. That a ∈ N implies ιiai ∈ N was shown earlier inthe proof. Since a ∈ N implies that all but finitely many coordinatesare 0, a is the sum of finitely many of the ιiai, proving the otherdirection. Hence

a ∈M(αN)⇔ aiαNzi for all i

⇔ ιid(ai, zi, zi),∈ N for all i

⇔ ιiai ∈ N for all i

⇔ a ∈ N.

This completes the proof. �

The behavior of the above isomorphism on finitely generated con-gruences and finitely generated submodules is developed in the exer-cises.

Corollary 9.10. Let β > 0 be an Abelian congruence on A andlet γ ∈ Con A satisfy β ≤ γ ≤ (0 : β). Let zi, i < λ, be elements ofA such that each γ–class contains at least one zi.

(i) If β is a minimal congruence, then∑

M(β, zi) is simple as anR(V, λ)–module.


(ii) If A is subdirectly irreducible, then∑

M(β, zi) is subdirectlyirreducible as an R(V, λ)–module.

Proposition 9.11. Suppose that [α, β] = 0 in Con A where α ≥β, and z α z′. Then the map x 7→ d(x, z, y′) is an R(V, 1, κ)–moduleisomorphism of M(β, z) onto M(β, z′) with inverse y 7→ d(y, z′, z).

Proof. This follows from straightforward calculations using Propo-sition 5.7 Let ψ be the map given in the proposition. To see that ψrespects the ring multiplication, let r(u, v,y) ∈ R(V, 1, κ) we use v inplace of v0) and x ∈M(β, z) and ck ∈ A, k < κ. Then

r · ψ(x) = r(d(x, z, z′), z′, . . . , ck, . . . )

= r(d(x, z, z′), d(z, z, z′), . . . , d(ck, ck, ck), . . . )

= d(r(x, z, c), r(z, z, c), r(z′, z′, c))

= d(r(x, z, c), z, z′)

= ψ(r · x)

We leave the rest of the proof to the reader. �

Now we begin our study of Abelian varieties. The next result showsthat the situation for Abelian varieties is much simpler than the generalcase. If R is a ring we let Mλ(R) denote the ring of all λ by λ matricesover R such that all but finitely many entries of each column are 0.

Theorem 9.12. Suppose that V = A is an Abelian variety. Then

(1) R(A, λ, κ) ∼= R(A, λ, 0)

(2) R(A, λ, κ) ∼= Mλ(R(A, 1, κ+ λ− 1))

Consequently, R(V, λ, κ) ∼= MλR(A, 1, 0).

Proof. Since A is Abelian, [θi, θi] = 0, and so Hij = Hij. As usuallet X = {u}∪{vi : i < λ}∪{yk : k < κ} and let X0 = {u}∪{vi : i < λ}.Recall that H0

ij , which we defined before Lemma 9.8, is defined just asHij except that we use FA(X0) instead of FA(X). There is an obviousembedding of H0

ij into Hij. To see that this embedding is onto letr(u,v,y) ∈ Hij and let s(u,v) = r(u,v,v0) where v0 is the vector ofall v0’s. (More formally s is the image of r under the endomorphismsending each yk to v0). Note s(u,v) ∈ H0

ij, since r(vi,v,y) = vj =s(vi,v) = r(vi,v,v0). Since the term condition holds universally inA, this last equation implies r(u,v,y) = r(u,v,v0) = s(u,v), showingH0ij = Hij . It is easy to see that these embeddings may be combined

to give a ring isomorphism which proves (1).


To see (2), fix i and consider the map ψkj : M(θi, vk) → M(θi, vj)given by ψkj(x) = d(x, vk, vj). Since [θi, 1] = 0, Proposition 9.11 impliesthat ψkj is a isomorphism with inverse ψjk. Now if r(u,v,y) ∈ Hij , thenr∗(u,v,y) = ψj0(r(ψ0i),v,y)) is in H00 since r∗(v0,v,y) = v0. Notethat we have the isomorphism H00

∼= R(A, 1, κ + λ − 1) by treatingvi, 1 ≤ i < λ, as “y variables”. Now if (rij) ∈ R(A, λ, κ), map (rij)to (r∗ij) ∈ Mλ(R(A, 1, κ + λ − 1)). Using Proposition 9.11, one caneasily verify that this map is one-to-one and onto. To see that it is ahomomorphism one needs to show that if r, t ∈ Hij , and s ∈ Hjk, then(r + t)∗ = r∗ + t∗, and (s · r)∗ = s∗ · r∗. Then, using Proposition 9.11,and suppressing all but the first variable we have

(r + t)∗ = (d(r(u), vj, t(u))∗

= ψj0d(r(ψ0iu), vj, t(ψ0iu))

= d(ψj0r(ψ0iu), ψj0vj, t(ψ0iu))

= d(ψj0r(ψ0iu), v0, ψi0t(ψ0iu))

= r∗ + t∗

and

(s · r)∗ = ψk0s(r(ψ0iu))

= ψk0s(ψ0jψj0r(ψ0iu))

= s∗ · r∗

This completes the proof. �

Theorem 9.12 shows that for an Abelian variety A only the ringR(A, 1, 0) is needed. We denote this ring R(A) and call it the ring ofA. The ideas developed above can be carried much further for Abelianvarieties, and we shall now discuss the Abelian case in detail. For therest of the chapter we will be working with a fixed Abelian variety A

and its ring R = R(A). Thus R = {r ∈ FA(u, v) : r(v, v) = v} withoperations

r + s = d(r, v, s)

−r = d(v, r, v)

r · s = r(u, v), v).

The variety of left R–modules is denoted M(R). If z ∈ A ∈ A andβ = 1A then we let M(A, z) denote M(1A, z). The action of R on A,given by (6), has the simple form

(8) r · a = r(a, z)


By the last proposition, the isomorphism type of this module is in-dependent of z. When the concrete representation of this module isimmaterial, we shall simply write it as M(A).

The correspondence between pairs (A, z) where z ∈ A ∈ A andmodules M ∈ M(R) where M = M(A, z), is actually onto M(R), aswe shall see. To completely specify the pair (A, z) we need, in general,not just the module M = M(A, z) but also a certain homomorphismof modules, ϕ : J→M where J is a module now to be defined. In factwe put J = M(FA(v), v), a module derived from the A–free algebra onone generator. For z ∈ A ∈ A, we write ϕ(z) for the A–homomorphismof FA(v) into A which carries v to z, (ϕ(z) is also a homomorphism ofmodules by formula (8).)

Here is how we can construct A, given M = M(A, z) and ϕ = ϕ(z).For an arbitrary n-ary term f(x1, . . . , xn) for A we define n elementsof R:

(9) t(j)(u, v) = d(t(v, . . . , v,ju, v, . . . , v), t(v, . . . , v), v)

where 1 ≤ j ≤ n. One can easily check the following, using the factthat every algebra in A is d-affine. For t an n-ary term operation,A ∈ A, and y1, . . . , yn, z ∈ A we have

(10) t(y1, . . . , yn) =

n∑

1

t(j) · yj + t(z, . . . , z)

(expressed with operations of M(A, z)).We can now construct an algebra A(M, ϕ) corresponding to any

M ∈ M(R) and ϕ ∈ Hom(J,M). We suppose that algebras in A arein the form a = 〈A, FA

i (i ∈ I)〉 with Fi ni-ary. Define bi ∈ J by

(11) bi = Fi(v, . . . , v).

For the universe of A(M, ϕ) we take the universe of M; and for i ∈ I,y1, . . . , yni

∈M we define

(12) Fi(y1, . . . , yni) =

ni∑

1

F(j)i · yj + ϕ(bi).

Then it is clear from (10), (11), (12) that, for z ∈ A ∈ A, we haveA = A(M(A, z), ϕ(z)). It is more difficult to verify that A(M, ϕ) ∈ A

for an arbitrary pair (M, ϕ). The proof begins with a lemma.

Lemma 9.13. (1) Let A, B ∈ A, ψ ∈ Hom(A,B) and ψ(z) =z′. Then ψ ∈ Hom(M(A, z),M(B, z′)).


(2) Let n ≥ 1 and FA(n+1) = FA(x1, . . . , xn, v). Then M(FA(n+1), v) = R · x1 ⊕ · · · ⊕R · xn ⊕ J and R · xi is the free cyclicR–module.

Proof. The first part follows immediately from formula (8) whichdefines the operations on the modules derived from A and B.

For the second part, we first note that any element ofM = M(FA(n+1), v) can be written as an A–term in x1, . . . , xnv, and by (10)

t(x1, . . . , xn, v) =n

∑

1

t(j) · xj + t(n+1) · v + t(v, . . . , v)

∈ R · x1 + · · ·+R · xn + J

since v = 0 in the module M (i.e., since t(n+1) · v = v). To see that thesum is direct and R · xi is free, suppose that

(∗)n

∑

1

rj · xj + s(v) = v

in M, with rj ∈ R, and s(v) ∈ J. Let e and ℓ be the endomorphismsof FA(n + 1) satisfying e(x1) = x1, e(x2) = . . . e(xn) = e(v) = v,ℓ(x1) = ℓ(v) = v. By (9,13)(1) e and ℓ are endomorphisms of M.Applying ℓ to (∗) we see that s(v) = v and applying e to (∗) givesr1 · x1 = v. This second relation says that A satisfies r1(x1, v) ≈ v, i.e.,r1 = 0 in R. Similarly, ri = 0, i = 2, . . . , n. �

Lemma 9.14. LetM ∈M(R) and ϕ ∈ Hom(J,M). ThenA(M, ϕ) ∈A.

Proof. We show that for any finite set {y1, . . . , ym} ⊆ A = A(M, ϕ),the subalgebra SgA({y1, . . . , ym} ∪ ϕ(J)) is a homomorphic image ofFA(m + 1) = FA(x1, . . . , xm, v). By Lemma 9.13 there is a homo-morphism of modules ψ : M(FA(m + 1), v) → M with ψ(xi) = yi,1 ≤ i ≤ m, and ψ|J = ϕ. We just have to check that ψ is also a homo-morphism of FA(m+1) into A. Let i ∈ I and u1, . . . , uni

∈ FA(m+1).Then using (10), (11), and then (12) we have

ψ(Fi(u1, . . . , uni)) = ψ(

ni∑

1

F(j)i · uj + Fi(v, . . . , v))

=

ni∑

i

F(j)i · ψ(Uj) + ψ(bi)

= F(A)i (ψ(u1), . . . , ψ(uni

))

since ψ(bi) = ϕ(bi). �


Lemma 9.15. Let M ∈M(R), ϕ ∈ Hom(J,M), and A = A(M, ϕ).Then M = M(A, 0) and ϕ = ϕ(0) : FA(v) → A where 0 is the zeroelement of M.

Proof. That M and M(A, 0) have the same operations as R-modules follows from (8), and from the fact that for any a, b, c ∈ A, thehomomorphism ψ : FA(x1, x2, x3, v) → A mapping x1, x2, x3 onto a,b, c and v onto 0 is a homomorphism of M(FA(4), v) into M, as shownin the proof of Lemma 9.14. Since ψ|FA(v) = ϕ, it follows ϕ = ϕ(0). �

Our proofs of the foregoing lemmas obscure the difficult point,which was to show that in A = A(M, ϕ), d(A)(a, b, c) = a − b + cevaluated in M. This is certainly not obvious, since d may be a com-plicated composition of the basic operations of A, but it is true byLemmas 9.14 and 9.15.

Let us now define M(R,J) to be the variety whose members areall the algebras B = (M, dj)j∈J in which M ∈ M(R) and ϕ(j) = djdefines a homomorphism of J into M. It is convenient to identify eachsuch algebra B with the corresponding ordered pair (M, ϕ). Recall

the definition of the ring elements F(j)i (for i ∈ I, 1 ≤ j ≤ ni). The

following theorem summarizes the facts proved above, and a little more.

Theorem 9.16. For an Abelian variety A, R = R(A), and J asdefined we have:

(1) A = {A(M, ϕ) : (M, ϕ) ∈M(R,J)},

(2) R is generated as a ring with identity by its elements {F (j)i },

and J is generated as an R–module by its element {bi}.(3) For z ∈ A ∈ A, we have that (M, ϕ) = (M(A, z), ϕ(z)) belongs

to M(R,J) and A(M, ϕ) = A.(4) For (M, ϕ) ∈ M(R,J) we have that A = A(M, ϕ) belongs to

A, and M = M(A, 0), and ϕ = ϕ(0).(5) For A = A(M, ϕ), A′ = A(M′, ϕ′) and any mapping ψ : A→

A′ of sets, ψ is an M(R,J)−−homomorphism if and only ifψ is an A–homomorphism and ψ(0) = 0′.

(6) FA(k + 1) ∼= A(R(k) ⊕ J, ι) with ι the inclusion map, for anycardinal k ≥ 0.

(7) A = V(FA(2)).

Proof. All parts except (2) and (7) are obvious from Lemma 9.13,Lemma 9.14, and Lemma 9.15 and their proofs. We leave (2) and (7) tothe reader. Note that in (7), what is to be proved is that any identitys(x1, . . . , xn) ≈ t(x1, . . . , xn) which fails to hold in A is refuted in somealgebra of A generated by two elements. The proof uses the affinerepresentations of s and t given by (10). �


Remark 9.17. The theorem can be viewed as stating the defi-nitional equivalence of the variety M(R,J) with the variety A(1) ofpointed A-algebras, i.e., algebras (A, z) where A ∈ A and z is an ar-bitrary element of A. The operations of the corresponding algebrasunder this definitional equivalence, (M, ϕ) and (A(M, ϕ), 0), are uni-formly interdefinable by terms, those of the one algebra from those ofthe other. We note that M(R) is essentially the class {(M, ϕ) : ϕ ≡ 0}and thus is contained in A, in a sense (see the comments below). A

is definitionally equivalent with M(R) if and only if A has a constantterm t(x), i.e., t(x) ≈ t(y) holds in A, whose value is an idempotentelement in every algebra of A.

We have seen that any Abelian variety A determines a system

σ(A) = (R,J, F(j)i , bj)i∈I,j≤ni

where {F(j)i } generates R as a ring and

{bi} generates J as an R-module. Conversely, A is completely deter-mined by σ(A) = σ, i.e., A = A(σ). It would probably be worthwhile tostudy the conditions under which an arbitrary system (R,J,F,b) = σwill determine a modular Abelian variety A(σ) via the definitions (12).[Note that A(σ) = {A(M, ϕ) : (M, ϕ) ∈ M(R,J)}, defined by (12),will not always be a variety.] We have not studied this question, al-though it is the natural first step in generating and classifying all pos-sible Abelian varieties.

We close this section with some observations that have been usefulin applications. Let A, R, and J be as in theorem 9.16. The zeromap ϕ(j) ≡ 0 of J into any module will be denoted by 0. Call analgebra A ∈ A linear if A = A(M, 0) for some M. This condition isobviously equivalent to: A has a one element subalgebra {e}, i.e., anidempotent element e (for then we can take M = M(A, e) and we haveϕ(e) = 0). The class of linear algebras in A need not be a variety. (Foran interesting case in which it is, see Theorem 12.4.)

There are two ways to construct a linear algebra from a givenA ∈ A, and they give the same result, up to isomorphism. Recallthe definition of ∆1,1 given in Chapter 4. We define the linearizationof A to be A∇ = A2/∆1,1.

Proposition 9.18. Let A ∈ A where A is an Abelian variety.Then

(i) For any z ∈ A, A∇∼= A(M(A, z), 0), thus it is linear.

(ii) A×A∇∼= A×A.

Proof. From the proof of Proposition 5.7 we have 〈x, y〉 ∆1,1 〈u, v〉if and only if v = d(y, x, u) if and only if v − u = y − x in one (equiva-lently all) of the modulesM(A, z). Now the map which sends 〈x, 〈y, z〉〉

EXERCISES 87

to 〈x, d(x, y, z)〉 is a homomorphism of A×A2 onto A2 and its kernelis precisely 0A ×∆1,1, hence we have (ii).

For (i) we have A = A(M(A, z), ϕ(z)) and so the map 〈x, y〉 7→d(y, x, z) is easily seen to be a homomorphism ofA2 ontoA(M(A, z), 0).The kernel of this homomorphism is ∆1,1. �

Proposition 9.19. Let z ∈ A ∈ A where A is an Abelian variety.

(i) Con A = Con(M(A, z)),(ii) If A is simple, then it has no proper subalgebras except one

element subalgebras.

Proof. A andM(A, z) have the same polynomial operations, hencethe same congruences. If S is any subalgebra of A and s0 ∈ S, then{〈x, y〉 : d(x, y, z0) ∈ S} = θ is a congruence of A and s0/θ = S. �

Exercises

1. Let (A, ·, /, \) be a quasigroup. Show that A satisfies the iden-tity (x · y) · (u · v) = (z · u) · (y · v) if and only if A is Abelianand R(V(A)) is commutative. For an arbitrary algebra Ain a modular variety show that A is Abelian and R(V(A))is commutative if and only if V(A) is permutable and everybasic operation of A commutes with every idempotent termoperation of A.

2. Show that if V is the variety of quasigroups then R(V) is notcommutative.

3. Let β > 0 inCon A, A ∈ A, and suppose that β is an Abeliancongruence. Let zi, i < λ, be a system of representatives forthe β-classes and let R = R(V, λ). By Corollary 9.10 M(β)is a simple R–module. Let D be the endomorphism ring ofM(β) as an R–module. Then show that D is a division ringand each M(β, zi) is a D-subspace of M(β). Moreover, showthat if a1, . . . , an is a finite set of D-linearly independent el-ements of M(β, zi) and b1, . . . , bn ∈ M(β, zj) then there is ag ∈ Hom(β, zi, zj) with g(ak) = bk, k = 1, . . . , n. If A is finiteshow there is a fixed prime p such that each β-block has orderpi for some i with i but not p depending on the block.

4. Suppose that α/β and γ/δ are projective quotients in Con A.Then the centralizer of α over β equals the centralizer of γ overδ, i.e.,

(β : α) = (δ : γ).


Suppose in addition that [α, α] ≤ β and [γ, γ] ≤ δ and letϕ = (β : α) = (δ : γ). Let zi, i < λ, be a system ofrepresentatives of the ϕ classes of A. Let M(α/β, zi) denoteMA/β(α/β, zi/β). Prove that

∑

i

M(α/β, zi) ∼=∑

i

M(γ/δ, zi)

as R(V, λ)–modules. This can be viewed as a version of thesecond isomorphism theorem.

5. Consider the situation of Theorem 9.9. Show that, underthe lattice isomorphism of β/0 onto the submodule lattice of∑

M(β, zi) given in the proof of that theorem, principal con-gruences contained in β map to cyclic submodules. Hencen-generated congruences are mapped to n-generated modules.

6. Let β be an Abelian congruence on A, γ = (0 : β), and letzi ∈ A, i < λ represent all the classes of γ. Let ck ∈ A, k < κ.Let X0 = {u}∪{vi : i < λ}, and X = X0∪{yk : k < κ}. Showthat the map σ : X → X0 which fixes the elements of X0 andmaps yk 7→ vi, where i is the first index with ck γ zi, induces aring homomorphism of R(V, λ, κ) onto R(V, λ) (= R(V, λ, 0)).Show that the kernel of this homomorphism is contained in thekernel of the action of R(V, λ, κ) on

∑

M(β, zi).

7. Let β be an Abelian congruence on A ∈ V and let γ = (0 : β).Suppose zi, z

′i, i ∈ I, are elements of A such that zi γ z

′i. Show

that∑

M(β, zi) and∑

M(β, z′i) are isomorphic asR(V, λ, κ)–modules, for any choice of constants ck ∈ A, k < κ.

CHAPTER 10

Structure and Representation

in Modular Varieties

We begin with some generalizations to modular varieties of Jonsson’stheorem for distributive varieties.

1. Birkhoff-Jonsson Type Theorems For Modular Varieties

Recall that the operators H , S, P, Ps, and Pu are the closuresunder homomorphic images, subalgebras, direct products, subdirectproducts, and ultraproducts, respectively. We understand these in theinclusive sense so that S(K), for instance, includes all isomorphic copiesof its members. Garrett Birkhoff’s theorem that HSP(K) is identicalwith the variety generated by K (when K is a class of similar alge-bras) was strengthened by Bjarni Jonsson [54] for the case that V (K)is distributive, to read V (K) = PsHSPu(K), i.e., every subdirectlyirreducible algebra in V (K) is a homomorphic image of a subalgebraof an ultraproduct of algebra in K. This theorem has had a profoundinfluence in shaping the themes and emphasis of subsequent researchon varieties.

Under the weaker assumption of modularity. Jonsson’s theorem isnot true. The following theorem, however, does generalize Jonsson’stheorem to modular varieties. It represents the combined work of Hage-mann, Herrmann, Freese, McKenzie, and Hrushovskii. Recall that themonolith of a subdirectly irreducible algebra is its least nontrivial con-gruence µ and that the centralizer (or annihilator) of µ is the largestcongruence α with [µ, α] = 0, i.e., α = (0 : µ). In particular if µ isnon-Abelian then α = 0.

Theorem 10.1. Suppose that V(K) is modular. If B ∈ V(K)is subdirectly irreducible and α is the centralizer of the monolith ofB, then B/α ∈ HSPu(K). Moreover, if V(K) is locally finite, thenB/γ ∈ SPuHS(K) for some γ ≤ α.

Note that ifV (K) is distributive then α = 0 and we obtain Jonsson’sTheorem. Also note that the theorem implies that either B has anAbelian monolith or B ∈ HSPu(K).

89

90 10. STRUCTURE AND REPRESENTATION

Proof. By Birkhoff’s theorem that V (K) = HSP(K), we mayassume that B = C/θ where C ⊆

∏

i∈I Ai, Ai ∈ K. For J ⊆ I,let ηJ be the kernel of the natural map from C to

∏

i∈J Ai. Notethat ηJ ∧ ηK = ηJ∪K and that J ⊆ K implies ηJ ≥ ηK . Since B issubdirectly irreducible, θ is uniquely covered by a congruence ψ. If welet ϕ = (θ : ψ), then B/α ∼= C/ϕ by Proposition 4.4 and Remark 4.6.

If β, γ ∈ Con C, then

β ∧ γ ≤ θ implies either

β ≤ θ or γ ≤ θ or both β ≤ ϕ and γ ≤ ϕ.

For suppose β ∧ γ ≤ θ, β � θ, and γ � θ. Then β ∨ θ ≥ ψ and hence[γ, ψ] ≤ [γ, β ∨ θ] = [γ, β] ∨ [γ, θ] ≤ (γ ∧ β) ∨ θ = θ. Hence γ ≤ ϕ.Similarly β ≤ ϕ.

Let F be a filter on I maximal with respect to the property J ∈ F

implies ηJ ≤ θ. Let U be an ultrafilter extending F. We claim thatJ ∈ U implies ηJ ≤ ϕ. This is clear if J ∈ F, so assume J ∈ U−F. Bythe maximality of F neither J nor its complement J ′ can be adjoinedto F. This implies that there is a K ∈ F such that ηJ∩K � θ andηJ ′∩K � θ. Now ηJ∩K ∧ ηJ ′∩K = ηK ≤ θ. Hence by what we haveshown above, ηJ∩K ≤ ϕ. But ηJ ≤ ηJ∩K , so the claim is proved.

Let ηU ∈ Con C be the restriction of the ultraproduct congruenceon

∏

I Ai to C. Then C/ηU is a subalgebra of the ultraproduct of∏

I Ai by U. It follows from the claim that ηU ≤ ϕ so that B/α ∼=C/ϕ ∈ HSPu(K).

In order to prove the last statement of the theorem we need a lemmashown to us by J. B. Nation.

Lemma 10.2. If B is a subdirectly irreducible algebra, then B is asubalgebra of an ultraproduct of a family Bi, i ∈ I, of finitely generated,subdirectly irreducible algebras from HS(B). Moreover, if µi is themonolith of Bi and µ is the ultraproduct congruence determined by theµi, then µ restricted to B is nonzero.

Proof. Let Cg (a, b) be the monolith of B. Let S be the collectionof all finite subsets of B which contain a and b. For S ∈ S let ψS ∈Con Sg(S) be a maximal congruence not containing 〈a, b〉. Let F bethe filter on S consisting of all T ⊆ S such that there is an S0 ∈ S

with {S : S ⊇ S0} ⊆ T. Let U be an ultrafilter extending F. LetD =

∏

SSg(S)/ψS. Let ϕ : B → D be given by ϕ(x)S = x/ψS if

x ∈ S and an arbitrary element of Sg(S)/ψS otherwise. Let D/U bethe ultraproduct of the Sg(S)/ψS corresponding to U. We leave itto the reader to check that the composite map ϕ : B → D/U is ahomomorphism. It is an embedding because ϕ(a) 6= ϕ(b). The last

1. BIRKHOFF-JONSSON TYPE THEOREMS 91

part of the lemma follows from the fact that 〈ϕ(a), ϕ(b)〉 is in theultraproduct congruence determined by the monoliths of the factors.

�

Returning to the proof of the theorem, suppose thatV (K) is locallyfinite and B ∈ V (K) is subdirectly irreducible. By the lemma, B is asubalgebra of an ultraproduct (

∏

Bi)/U of finite subdirectly irreduciblealgebras Bi in V (K). Let αi be the centralizer of the monolith µi ofBi. By the first part of the theorem, Bi/αi ∈ HSPu(K). Since Bi isfinite and V (K) is locally finite, this implies Bi/αi ∈ HS(K), becauseBi is a homomorphic image of a finite free algebra in V (K) and thusthere is a finite subset K1 ⊆ S(K) of finite algebras with B1 ∈ V (K1).

Let α = (Παi)/U and µ = (Πµi)/U ∈ Con(ΠBi/U) be the cor-responding ultraproduct congruences. Using our syntactic definitionDefinition 3.2 of the commutator, it is easy to see that [α, µ] = 0 inCon(ΠBi/U). Let γ be the restriction of α and β be the restriction ofµ to B. By the lemma β > 0 and so β ≥ µ. By Proposition 4.4(2),[β, γ] = 0 in Con B. Thus γ ≤ α. Moreover, B/γ ∈ SPuHS(K). �

For distributive, locally finite varieties Theorem 10.1 gives a strength-ening of Jonsson’s theorem. Explicitly we have:

Corollary 10.3. Let V(K) be distributive and locally finite. IfA ∈ V(K) is subdirectly irreducible, then A ∈ SPuHS(K).

Remark 10.4. (1) Let A be finite and let B ∈ V (A) be sub-directly irreducible with monolith µ and α = (0 : µ) its cen-tralizer. Theorem 10.1 gives the following. If |B| > |A|, thenα ≥ µ and by Proposition 9.11, there are at most |A| isomor-phism types of modules among the M(µ, x), x ∈ B. Also, eachof these modules has size at most |A| by Theorem 10.16 tofollow. This situation will be investigated more thoroughly inthe next section.

(2) The following example shows that B/γ ∈ SPuHS(K) requireslocal finiteness (or some similar assumption), in both Theo-rem 10.1 and Corollary 10.3. Take K to be all finite semidis-tributive lattices. These include all splitting lattices and soby a result of A. Day [21], V (K) is the variety of all lattices.Hence by Jonsson’s Lemma HSPu(K) contains all subdirectlyirreducible lattices. Clearly S(K) = K and since the membersofK are finite, H(K) = K also. Thus SPuHS(K) = SPu(K).The latter class is clearly contained in the class of semidis-tributive lattices. But obviously there are subdirectly lattices


which are not semidistributive. Hence SPuHS(K) is properlycontained in HSPu(K).

The next theorem is a nice application of Theorem 10.1.

Theorem 10.5. Suppose that B ∈ V(K), where V(K) is modular.If Con B has finite length (in particular if B is finite) then B has anilpotent congruence θ such that B/θ is a subdirect product of algebrasin HSPu(K).

Proof. Let ϕ =∧

β<α(β : α) be the intersection of all the central-izers of the prime quotients of Con B. If 0 6= γ ∈ Con B is arbitrarythen since Con B has finite length, there is a δ < γ. By definition[ϕ, γ] ≤ δ. Hence for any γ 6= 0, [ϕ, γ] < γ. It follows immediately fromthis and the finite length of Con B that ϕ is nilpotent. (In fact it iseasy to see that ϕ is the largest nilpotent congruence, cf. Exercise 6.4.)Now let β < α and let δ be a completely meet irreducible congruencewith δ ≥ β but δ � α. Let γ = α ∨ δ . Then α/β ր γ/δ. By Exer-cise 9.4 (β : α) = (δ : γ) and by Theorem 10.1 B/(δ : γ) ∈ HSPu(K).Since ϕ is the intersection of such congruences, the theorem follows. �

As we mentioned above, ϕ is the largest nilpotent congruence. Itis called the Fitting congruence in analogy to group theory. Noticethat even without the assumption that Con B has finite length weobtain a congruence ϕ with B/ϕ a subdirect product of algebras inHSPu(K) and [ϕ, γ] ≤ δ whenever γ > δ. You are asked to prove aslightly stronger result in Exercise 1.

2. Subdirectly Irreducible Algebras inFinitely Generated Varieties

In this section we strengthen the results of the last section in thecase V = V (A), A finite. We begin with the concept of similarity.

Let θ be an Abelian congruence on an algebra A ∈ V and let α beits centralizer, i.e., α = (0 : θ) . Let z1, i < λ, be a set of distinctrepresentatives of the α–classes. Recall the definition of the matrixring R(V, λ) = R(V, λ, 0) (Definition 9.3) and of the modules M(θ, zi).As pointed out in Chapter 9,

∑

i<λM(θ, zi) is an R(V, λ)–module.

Definition 10.6. Let θ ∈ Con A and ψ ∈ Con B be Abeliancongruences with centralizers α and β, respectively. We say that θ inA is similar to ψ in B if the following two conditions hold. Firstthe is an isomorphism σ : A/α → B/β. Let zi, and wi, i < λ, be asystems of distinct representatives of the α–classes of A and the β–classes of B, respectively, indexed so that σ(zi/α) = wi/β. The second

2. SUBDIRECTLY IRREDUCIBLE ALGEBRAS 93

condition is that∑

i<λM(θ, zi) and∑

i<λM(ψ,wi) are isomorphic asR(V, λ)–modules.

Definition 10.7. Subdirectly irreducible algebras A and B andsaid to be similar if they are isomorphic or both have Abelian mono-liths and µA in A is similar to µB in B. We denote this by A ∼ B.

Notice that if A and B are similar with Abelian monoliths, thenthe Abelian groups determined by the blocks of the respective mono-liths are isomorphic if these blocks lie in corresponding blocks of thecentralizers. This follows easily from the fact that

∑

M(θ, zi) is iso-morphic to

∑

M(ψ,wi). The main result of this section shows that ifB ∈ V (A) is subdirectly irreducible and A is finite, then B is simi-lar to an algebra in HS(A). Notice that this is considerably strongerthan Theorem 10.1.

The next theorem gives a useful lattice-theoretic characterizationof similarity.

Theorem 10.8. Let A and B be subdirectly irreducible algebras ina modular variety V. Then A ∼ B if and only if there is an algebraC ∈ V and γ, δ, η, ǫ ∈ Con C, η and ǫ completely meet irreducible(with upper covers η∗ and ǫ∗, such that A ∼= C/η, B ∼= C/ǫ, and

η∗/η ց γ/δ ր ǫ∗/ǫ.

Moreover, if A ∼ B then such a C can be taken to be a subdirectproduct of A and B with η and ǫ as the projection kernels.

Proof. Let A and B have monoliths µ and ν with centralizers αand β, respectively. Suppose C and γ, δ, η, and ǫ are as describedin the theorem. That A is similar to B is precisely the content ofExercise 4.

Now assume A ∼ B. The case A ∼= B is easy so we assume µand ν are Abelian and thus α ≥ µ and β ≥ ν. Let zi, wi, i < λ,be as in the definition of similarity. Let τ be a R(V, λ) isomorphismof

∑

i<λM(µ, zi) onto∑

i<λM(ν, wi) and let τi be its restriction toM(µ, zi). Since τ is a R(V, λ) isomorphism, τi is an R(V, 1) isomor-phism of M(µ, zi) onto M(ν, wi).

Let D = A/λ and let h be the natural map of A onto D and letk : B → D with ker k = β. We may assume that h(zi) = k(wi). LetC be the subalgebra of A × B with universe C = {(x, y) ∈ A × B :h(x) = k(y)}.

Define γ ∈ ConC by (a, b) γ (a′, b′) if a µ a′, b ν b′, and τi(d(a′, a, zi)) =

d(b′, b, wi) where i is the index with a λ zi. (Note that b β wi sincek(b) = h(a) by the definition of C, h(a) = h(zi) as a α zi and α is the


kernel of h, and h(zi) = k(wi) by our assumption. Thus k(b) = k(wi)and, since β is the kernel of k, b β wi.) We view d(a′, a, zi) as the ele-ment of

∑

M(µ, zi) whose ith coordinate is d(a′, a, zi) and whose other

coordinates are 0. To see that γ is a congruence is just a matter ofchecking a few details. To see symmetry, for example, note that sinceτ is a module homomorphism τ(−x) = −τ(x), i.e., τ(d(zi, x, zi)) =d(wi, τ(x), wi), for x ∈ M(µ, zi). Now suppose (a, b) γ (a′, b′) so thatτ(d(a′, a, zi)) = d(b′, b, wi). By Proposition 5.7

d(zi, d(a′, a, zi), zi) = d(d(a, a, zi), d(a

′, a, zi), d(a′, a′, zi))

= d(d(a, a′, a′), d(a, a, a′), d(zi, zi, zi))

= d(a, a′, zi).

Hence τ(d(a, a′, zi)) = d(b, b′, wi).Let η, ε ∈ ConC be the kernels of the projections and let η∗ and

ε∗ be their unique upper covers. We will show that η ∧ γ = ε ∧ γ = 0.Indeed, if (a, b) η ∧ γ (a′, b′), then a = a′ since η is the kernel of thefirst projection, and so by the definition of γ, wi = τi = τi(zi) =τi(d(a, a, zi)) = d(b′, b, wi). Hence d(b, b, wi) = wi = d(b, b, wi). Fromthis it follows that b = b′ (which follows from the term condition andthe fact that b′ ν b β wi)). Thus η ∧ γ = 0. Clearly γ ≤ η∗ ≻ η as〈a, b〉 γ 〈a′, b′〉 implies a µ a′ by definition. Now γ 6= 0. In fact if a,a′ ∈ M(µ, zi) and b ∈ M(ν, wi) then there is a b′ with 〈a, b〉 γ 〈a′, b′〉.Indeed, since τ is a R(V, λ) isomorphism, τ restricted to M(µ, zi) mapsM(µ, zi) onto M(ν, wi). Now the claim follows from Proposition 9.11(or directly from Proposition 5.7). It follows that η∨γ = η∗, and henceη∗/η ց γ/0ր η∗/ε. �

Definition 10.9. Let A be subdirectly irreducible with monolithµ and centralizer α = (0 : µ). If µ is non-Abelian let D(A) = A.Otherwise let D(A) = A(µ)/∆µ,α′. (Recall that A(µ) = {〈x, y〉 ∈A× A : x µ y} and that ∆µ,α is the congruence on A(µ) generated by{〈〈x, x〉, 〈y, y〉〉 : x α y}).

Part of Con A(µ) is shown in Figure 1.In the case µA is Abelian, D(A) has two nice properties. First,

the next theorem will show that in D(A), µ = α. Secondly, D(A)has a subalgebra which is a transversal for its monolith. Namely,{〈x, x〉/∆µ,α : x ∈ A} is such a subalgebra.

Theorem 10.10. Suppose that A is subdirectly irreducible. ThenD(A) is subdirectly irreducible and A ∼ D(A). Moreover if the mono-lith of A is Abelian then the monolith of D(A) is its own centralizer and

2. SUBDIRECTLY IRREDUCIBLE ALGEBRAS 95

η1

∆

α0 = α1

η0

µ = µ1

Figure 1.

{〈x, x〉/∆µ,α : x ∈ A} is a subalgebra of D(A) which is a transversalfor the monolith of D(A).

Proof. Recall that if θ ∈ Con A then θi ∈ Con A(µ), i = 0, 1, isdefined by 〈a0, a1〉 θi 〈b0, b1〉 if ai θ bi. Let ηi, i = 0, 1, be the projectionkernel of the projection map, i.e., ηi = 0i. Since the theorem is obviousif µ is non-Abelian, we assume µ is Abelian and let α = (0 : µ). Write∆ for ∆µ,α. We will show that the following transposition holds inCon A(µ) (see Figure 1).

(*) α0/∆ց η1/0ր µ0/η0

First it is not hard to see that µ0 = µ1, α0 = α1, and η0∨η1 = µ0 = µ1.Indeed, if 〈a0, a1〉 α0 〈b0, b1〉 in A(µ) then, by the definition of A(µ)and of α0, a1 µ a0 α b0 µ b1. Since µ ≤ α, 〈a0, a1〉 α1 〈b0, b1〉. The proofof the first equality is the same. To see the third equality suppose that〈a0, a1〉 µ0 〈b0, b1〉. Then all four elements are in the same µ–class andso 〈a0, a1〉 η0 〈a0, b1〉 η1 〈b0, b1〉. Since the other inclusion is trivial, thisproves the third equality. Now if 〈a0, a1〉 α0 〈b0, b1〉 then a0 α b0 andso 〈a0, a1〉 η0 〈a0, a0〉 ∆ 〈b0, b0〉 η0 〈b0, b1〉. Hence α0 = α1 = ∆ ∨ η0,and similarly α0 = α1 = ∆∨ η1. If 〈a0, a1〉 ∆ ∧ η1 〈b0, b1〉 then a1 = b1.Now by Theorem 4.9(iv), a0 [α, ν] b0, i.e., a0 = b0 and so ∆ ∧ η1 = 0.Thus (∗) holds. By Theorem 10.8 this will show that A ∼ D(A) oncewe have shown that D(A) is subdirectly irreducible. To do this we willshow that ∆ is uniquely covered by α0 = α1.

Suppose ϕ > ∆ in Con A(µ). If ϕ ≥ η1 then ϕ ≥ η1 ∨ ∆ = α0.If ϕ 6≥ η1 then ϕ ∧ η1 = 0 as η1 ≻ 0 (which follows from the the factthat η1/0 ր µ0/η0 and η0 ≺ µ0). Now [ϕ ∨ η0, µ0] = [ϕ ∨ η0, η0 ∨η1] ≤ η0 ∨ (ϕ ∧ η1) = η0. It now follows from the definition of α thatϕ ≤ ϕ∨ η0 ≤ α0. Since α0 > ∆ by (∗), we must have ϕ = α0, contraryto ϕ 6≥ η1. Thus ∆ is completely meet irreducible and D(A) is thereforesubdirectly irreducible.

To show that the monolith of D(A) = A(µ)/∆ it its own centralizerwe need to show that (∆ : α0) = α0 in Con A(µ). Clearly (by (∗)


for example) α0 ≤ (∆ : α0). To see the other inclusion, suppose thatϕ ∈ Con A(µ) satisfies [ϕ, α0] ≤ ∆. Then [ϕ, η0] ≤ [ϕ, α0] ≤ ∆.So [ϕ, η0] ≤ ∆ ∧ η0 = 0. Similarly [ϕ, η1] = 0 and thus [ϕ, µ0] =[ϕ ∧ η0 ∨ η1] = 0. Since α0 = (η0 : µ0), ϕ ≤ α0, as desired.

For the last statement of the theorem let S = {〈x, x〉/∆µ,α : x ∈ A}.Clearly S is a subalgebra of D(A) = A(µ)/∆. To see that it is atransversal of the monolith of D(A) it suffices to show that if 〈x, x〉 α0

〈y, y〉 then 〈x, x〉 ∆ 〈y, y〉 and if 〈x, y〉 ∈ A(µ) then 〈x, y〉 α0 〈y, y〉.By its definition ∆ = ∆µ,α is the congruence of A(µ) generated bythe pairs 〈〈x, x〉, 〈y, y〉〉 such that x α y and hence the first implicationabove holds. Since µ ≤ α, the second implication is trivial. �

Theorem 10.11. Suppose that A and B are subdirectly irreducible.Then A ∼ B if and only if D(A) ∼= D(A).

Proof. If D(A) ∼= D(A) then by the previous theorem A ∼D(A) ∼ D(A) ∼ B. For the other direction, assume A ∼ B. Ifthey have non-Abelian monoliths then D(A) ∼= D(A), trivially. So weassume the monoliths are Abelian. Now D(A) ∼ A ∼ B ∼ D(A); soit suffices to show that if A and B are similar algebras, each with itscentralizer equal to its monolith and each having a subalgebra whichis a transversal for its monolith, then A ∼= B.

Since A ∼ B we have an isomorphism τ of∑

i<λM(µ, zi) onto∑

i<λM(ν, wi) where of course µ is the monolith of A, ν is the mono-lith of B, zi, i < λ, is a transversal for α = (0 : µ), and wi, i < λ,are a transversal for ν. As pointed out earlier, τ maps M(µ, zi) iso-morphically onto M(ν, wi). Since µ = (0 : µ), A is the disjoint unionof the M(µ, zi)’s and similarly B is the union of the M(β, wi)’s. Thusτ defines a bijection from A to B. We shall show that τ is a ho-momorphism. By Exercise 7 the isomorphism type of

∑

i<λM(µ, zi)does not change if we replace zi with an element which is in the sameµ–class. Now A has a subalgebra which is a transversal for µ which,by these remarks, we may assume is {zi : i < λ} and similarly, wemay assume that {wi : i < λ} is a subalgebra of B. Let f be a termand x1, . . . , xm ∈ A. Let x0 = f(x1, . . . , xn) and suppose xt ∈ zt/µ,t = 0, 1, . . . , n. Since f(z1, . . . , zn) µ f(x1, . . . , xn) = x0 µ z0, and thezi’s form a subalgebra, f(z1, . . . , zn) = z0. Now an easy application ofProposition 5.7 shows

f(x1, . . . , xn) = f(x1, z2, . . . , zn) + . . .+ f(z1, . . . , zm−1, xn).

The addition hare is in M(µ, z0) where z0 is the zero element. Since τpreserves addition, it suffices to show that

τf(x1, z2, . . . , zn) = f(τx1, w2, . . . , wn).

3. RESIDUALLY SMALL VARIETIES 97

This follows easily from the fact that τ is a R(V, λ)–homomorphism(see the comment after Definition 9.3. �

Theorem 10.12. If B is a subdirectly irreducible algebra in V(A),where A is finite, then B is similar to a subdirectly irreducible algebrain HS(A).

Proof. By the last theorem it suffices to prove this for D(A). Sowe may assume that B = D(A). Let µ be the monolith of B. If(0 : µ) = 0, then B ∈ HS(A) by Theorem 10.1. So we may assumethat (0 : µ) = µ. By Theorem 10.1, B/µ ∈ HS(A). Theorem 10.16below shows that each block of µ has size at most |A|. Hence |B| ≤|A|2. In particular, B is finite. If follows that B = C/θ for someθ ∈ Con C, where C is a subdirect product of subdirectly irreduciblealgebras A0, . . . ,Ak−1 ∈ HS(A). Let η0, . . . , ηk−1 ∈ Con C be theprojection kernels. For subsets {ai} and {bj} of a lattice let {ai} ≫ {bj}mean that for each i there is a j with ai ≥ bj . Since C is finite, we maychoose θ0, . . . , θm−1 ∈ Con C maximal in the ordering ≫ such that∧

θi ≤ θ and {θi} ≫ {ηj}. By relabeling these θi’s as ηi’s (this amountsto a new choice of the Ai’s, but by the second condition above they arestill in HS(A) and of course subdirectly irreducible) we may assumethat if ϕ0, . . . , ϕm−1 ∈ Con C satisfy (1)

∧

ϕi ≤ θ, and (2) for eachi < m there is a j with ϕi ≥ ηj, then {η0, . . . , ηn−1} ⊆ {ϕ0, . . . , ϕm−1}.If θ ∧ η′i > 0 we could replace ηi with ηi ∨ (θ ∧ η′i) and violate theabove conditions. So θ ∧ η′i = 0. θ ≥ η′i would also violate the aboveconditions. Thus θ∨η′i ≥ θ∗, where θ∗ is the unique congruence coveringθ. It follows that θ∗∨η′i = θ∨η′i. Modularity now yields that θ∗∧η′i > 0.Hence θ∗/θ ց θ∗∧η′i/0. Since ηi∧η

′i = 0 and θ∗∧η′i > 0, ηi∨(θ

∗∧η′i) =η∗i . Thus

θ∗/θց θ∗ ∧ η′i/0ր η∗i /ηi.

Hence by Theorem 10.8 B ∼ Ai, i = 0, . . . , k − 1. �

We record the fact the similarity types cannot mix in the nextcorollary.

Corollary 10.13. If B ∈ V(A) is subdirectly irreducible, A andB are finite, then for some n, B ∈ V(A1, . . . ,An) with Ai ∈ HS(A)and B ∼ Ai, i = 1, . . . , n. �

3. Residually Small Varieties

In this section we continue our study of subdirectly irreducible al-gebras in a finitely generated variety. A variety is residually small if ithas only a set of isomorphism types of subdirectly irreducible algebras.


Equivalently, there is a cardinal κ bounding the cardinality of all thesubdirectly irreducible algebras in the variety. Quackenbush’s problemasks the following question. If a finitely generated variety contains noinfinite subdirectly irreducible algebras is there an integer n boundingthe size of all subdirectly irreducible algebras in the variety? In thissection we show that this is true for modular varieties. In fact we willshow more. Namely we will show that any finitely generated, resid-ually small, modular variety has a finite bound n for its subdirectlyirreducible algebras. We will also show that a finitely generated vari-ety is residually small if and only if it satisfies the congruence identity(C1) of Chapter 8. In fact any residually small modular variety satis-fies (C1). The next theorem pertains to any modular variety, not justfinitely generated varieties.

Theorem 10.14. Let V we a modular variety containing an algebraA with congruences β and γ satisfying

β ≤ [γ, γ] and [β, γ] < β.

Then V is not residually small.

Proof. We will actually prove more. We will show that if λ is anycardinal the is a subdirectly irreducible algebra in V whose congruencelattice has cardinality at least λ. Let A ∈ V satisfy the hypothesesof the theorem. Choose θ a completely meet irreducible congruencesuch that θ ≥ [β, γ] and θ 6≥ β. Let θ∗ be the unique cover of θ.Then it is easy to see that θ∗ ≤ [θ ∨ γ, θ ∨ γ] ∨ θ and [θ∗, θ ∨ γ] ≤ θ,since θ∗ ≤ θ ∨ β. Hence, by changing notation, we may assume thatA is subdirectly irreducible and that β is its monolith. Recall thatalgA(γ) is γ thought of as a subalgebra of algA× algA. Let κ = ∆γ,β,where ∆γ,β is the congruence on A(γ) defined in Definition 4.7. Letη0, η1 ∈ Con A(γ) be the kernels of the two projections onto A. Fromthe definition of ∆γ,β we easily see that κ ∧ η0 = κ ∧ η1 = 0 and thatκ ∨ ηi = βi, i = 0, 1. See Figure 2.

Let ℵ be an arbitrary cardinal and let B = {(aδ) ∈ Aℵ : aδ γaǫ for all δ, ǫ < ℵ}. Recall that if ψ ∈ Con A then ψǫ ∈ Con B isdefined by (aδ) ψǫ (bδ) if and only if aǫ ψ bǫ. Note that γδ = γǫ for allǫ, δ < ℵ. We again denote this congruence by γ. Easy calculations onthe elements show that ηǫ ∨ ηδ = γ for ǫ 6= δ. Let κδ ∈ Con B be

{〈(aǫ), (bǫ)〉 : 〈a0, aδ〉 ∆γ,β 〈b0, bδ〉 and aǫ = bǫ, ǫ 6= 0, δ}.

i.e., all pairs of elements of B equal in all coordinates except 0 and δand such that the image of the projection onto A ×A determined by


η1

κ

ρ1ρ0

η0

γ0 = γ1

Figure 2.

the 0th and δ coordinate is in ∆γ,β. Define

θδ = {〈(aǫ), (bǫ)〉 : aδ β bδ and aǫ = bǫ, ǫ 6= δ}.

Let η′δ =∧

ǫ 6=δ ηǫ. Finally set θ =∨

δ θδ and κ =∨

δ>0 κδ. Note

θ0 ≤ θδ ∨ κδ for δ 6= 0. To see this let (aǫ) θ0 (bǫ) so that a0 β b0 andall other coordinates are equal. Now by making changes only in the 0and δ coordinates, and using the fact that 〈a0, a0〉 ∆γ,β 〈b0, b0〉 by itsdefinition, we have

(a0, . . . , aδ, . . .) η′δ (a0, . . . , a0, . . .)

κδ (b0, . . . , b0, . . .)

η′δ (b0, . . . , bδ, . . .)

Thus θ0 ≤ η′δ∨κδ and meeting with βδ gives θ0 ≤ θδ∨κδ by modularity.A similar argument shows that θδ ≤ θ0 ∨ κδ.

Hence for δ 6= 0 6= ǫ we have

κ ∨ θδ = κ ∨ κδ ∨ κǫ ∨ θδ

= κ ∨ κǫ ∨ θ0 ∨ θδ

= κ ∨ θǫ ∨ θ0 ∨ θδ

≥ θǫ ∨ θ0

It follows that κ ∨ θδ = θ for all δ and this holds for δ = 0 as well.We also claim that θδ 6≥ κ. To see this first note that since β > 0

in Con A, θδ > 0 in Con B, and is thus compact. Hence if θδ ≤ κ,then θδ ≤ κǫ1 ∨ · · · ∨ κǫn, for some ǫ1, . . . , ǫn. First consider the caseδ = 0 and induct on n. Observe that θ0∧κǫ = 0 since if 〈(aδ), (bδ)〉 is inthis intersection then aδ = bδ for all δ 6= 0, and 〈a0, aǫ〉 ∆γ,β 〈b0, bǫ〉 =〈b0, aǫ〉. By Theorem 4.9 this implies that 〈a0, b0〉 ∈ [γ, β] = 0. Thisproves that n > 1. Arguing on elements it is easy to see that the θδ’s


are independent. Using this, and the fact that κǫ1 ≤ θ0 ∨ θǫ1 , and thatκǫ2 ∨ · · · ∨ κǫn ≤ θ0 ∨ θǫ2 ∨ · · · ∨ θǫn , and the induction hypothesis wehave

θ0 = θ0 ∧ (κǫ1 ∨ · · · ∨ κǫn)

= θ0 ∧ (θ0 ∨ θǫ1) ∧ (κǫ1 ∨ · · · ∨ κǫn)

= θ0 ∧ (κǫ1 ∨ [(θ0 ∨ θǫ1) ∧ (θ0 ∨ · · · ∨ θǫn) ∧ (κǫ2 ∨ · · · ∨ κǫn)])

= θ0 ∧ (κǫ1 ∨ (θ0 ∧ (κǫ2 ∨ . . . ∨ κǫn)))

= θ0 ∧ κǫ1 = 0

This contradiction shows θ0 6≥ κ. The case δ 6= 0 follows from the factthat κ ∨ θδ = κ ∨ θ0.

If follows that κ < θ. Let λ ∈ Con B be a maximal congruencewhich contains κ but not θ. Then λ is completely meet irreducible, i.e.,B/λ is subdirectly irreducible. Now in Con B, γ/ηδ ց η′δ/0. SinceA is subdirectly irreducible with monolith β, the interval γ/ηδ has aunique atom βδ and the interval η′δ/0 has a unique atom θδ. Henceλ ∧ η′δ = 0, for otherwise λ ≥ θδ, which implies λ ≥ θδ ∨ κ = θ,a contradiction. Thus ηδ ∨ η

′δ = γ and λ ∧ η′δ = 0. If for some δ,

λ ∨ ηδ ≥ γ, then [γ, γ] ≤ [ηδ ∨ η′δ, ηδ ∨ λ] ≤ ηδ ∨ (λ ∧ η′δ) = ηδ. By

Proposition 4.4 this would imply that in Con A, [γ, γ] = 0, contraryto hypothesis. Thus for each δ, λ∨ ηδ � γ. Since ηδ ∨ ηǫ = γ for δ 6= ǫ,the λ ∨ ηδ, δ < ℵ, must be pairwise distinct. Hence |Con B/λ| ≥ ℵ,proving the theorem. �

Theorem 10.15. Let A be an algebra in a modular variety and let|A| = m < ω. Then the following are equivalent.

(1) V(A) is residually small.

(2) V(A) is residually ≤ m+ ℓ ! where ℓ = mmm+3

.(3) For any µ, ν ∈ Con C, where C ⊆ A, ν ≤ [µ, µ] implies

ν = [µ, ν].

Proof. Trivially (2) implies (1) and Theorem 10.14 shows that(1) implies (3). So assume that (3) holds. It follows directly fromLemma 10.2 that if V (A) contained an infinite subdirectly irreduciblealgebra, it would have arbitrarily large finite subdirectly irreduciblealgebras (cf. Quackenbush [75]). Thus we only need to show thatevery finite subdirectly irreducible algebra in V (A) has cardinalitybounded as stated in (2). By Theorem 8.1, the condition of (3) holdsin every finite algebra in V (A).

Let B ∈ V (A) be a finite subdirectly irreducible algebra withmonolith β. If β is not Abelian then B ∈ HS(A) by Theorem 10.1and B is bounded as in (2). Thus we assume that β is Abelian and let


γ = (0 : β) be the centralizer of β. By Theorem 10.1 B/γ ∈ HS(A). If[γ, γ] 6= 0, then [γ, γ] ≥ β, and since γ is the centralizer of β, [β, γ] = 0.This however violates (3) so we conclude that [γ, γ] = 0. Thus γ isan Abelian congruence with, say, k blocks where k ≤ m = |A|. Letz1, . . . , zk be a system of distinct representatives for these blocks. Thenby Corollary 9.10, M(γ) =

∑ki=1M(γ, zi) is a subdirectly irreducible

R(V (A), k) = R(V (A), k, 0)–module.Now

|R(V (A), k, 0)| ≤[

mmk+1]k2≤ mmk+3

≤ mmm+3

.

These bounds follow from the fact that R(V (A), k) consists of k by kmatrices whose (i, j)th elements lie inHij which has cardinality at most|FV (A)(k + 1)|. Thus M(γ) is a subdirectly irreducible module over aring with cardinality bounded in terms of m = |A| as above. Thus wehave reduced our problem to the special classical case of bounding thesize of subdirectly irreducible modules over a finite ring in terms of thesize of the ring. (The variety of left (unitary) R–modules is generatedby R as a left R–module and thus is finitely generated when R is finite.A bound on M(γ) easily gives a bound on B, see below.) At this pointwe could use the classical results of Rosenberg and Zelinsky [78] toobtain a bound for M(γ). Instead we will give a direct argument.

Thus suppose that M is a subdirectly irreducible module over afinite ring R. Let N be the unique minimal submodule of M, and leta ∈ N, a 6= 0. Then, if c 6= 0 is in M, N ⊆ Rc. Hence for each c 6= 0in M we can assign an r ∈ R such that rc = a. This implies that|M | ≤ |R|!. We can prove this by induction as follows. We prove thatif we have Abelian groups K and L and an element a 6= 0 in L andk homomorphisms r1, . . . , rk ∈ Hom(K,L) such that for each c 6= 0 inK there is an ri with ric = a, then |K| ≤ (k + 1)!. We induct on k;the initial case is left to the reader. Now each nonzero element of K isassigned to some ri. Suppose that r1 has the most elements assignedto it, say c1, . . . , ct. Then |K| ≤ 1 + kt. Now let K1 be the kernel ofr1. Since c1− ci, i = 1, . . . , t, is in K1, |K1| ≥ t. Thus |K| ≤ 1+ k|K1|.Now K1 satisfies the same hypothesis as K using r2, . . . , rk. Thus byinduction |K1| ≤ k!, and |K| ≤ 1+ k · k! ≤ (k+1)!, proving the claim.

Hence we conclude that |M(γ)| ≤ ℓ !, where ℓ = mm+3

. Now M(γ)

is the direct product of the γ–blocks of B, so |M(γ)| =∏k

i=1 |M(γ, zi)|.

Clearly |B| =∑k

i=1 |M(γ, zi)|. From this it follows that |B| ≤ m +|M(γ)| ≤ m+ ℓ !, as desired. �


4. Chief Factors and Simple Algebras

Every compactly generated (i.e., algebraic) lattice is weakly atomic;that is, if θ < ψ in Con A the there are α, β ∈ Con A with θ ≤β ≺ α ≤ ψ (α covers β). In our paper [29] we defined ♯α/β to be thesupremum of the cardinalities of the sets {y/β : y ∈ x/α} for x ∈ A,i.e., ♯α/β is the supremum of the number of β–blocks lying in a singleα–block. These prime intervals α/β and their associated numbers ♯α/βwere loosely construed as a general analogue of what, in group theory,are called the chief factors of a group. [There are some open problemshere, It would be nice to have some canonical algebra associated withα, β with properties similar to chief factors in groups. A candidate forsuch an algebra in the case α/β is Abelian isM =

∑

M(α/β, zi), wherethe zi, s represent the classes of the congruence (β : α), (M(α/β, zi) isdefined in Exercise 4). By Definition 9.3, M is a simple module. SeeExercise 3 for additional properties of M.]

We proved some theorems in [29] about these “chief factors” offinite algebras in a modular variety generated by finite algebra. Theseresults extended similar results proved by J. B. Nation and WalterTaylor for permutable varieties. Here we improve these results.

Theorem 10.16. Let A be a finite algebra in a modular variety andlet B ∈ V(A).

(1) If α ≻ β in Con B, then ♯α/β ≤ |A|.(2) If the length of Con B is n, then |B| ≤ |A|n.(3) If B is simple, then |B| ≤ |A|.

Proof. The following facts are elementary and are left as exercisesto the reader (or see Freese-McKenzie [29]): (i) If γ ≤ β ≺ α then ♯α/βis the same in A as in A/γ. (ii) If α/β ց γ/δ then ♯α/β ≥ ♯γ/δ andequality obtains if β and γ permute.

To prove 1. we first observe that we may assume that B is subdi-rectly irreducible, β = 0, and α is the monolith. Indeed let λ be acompletely meet irreducible congruence with λ ≥ β and λ � α. Thenλ ∨ α/λ ց α/β and so by (i) and (ii) it suffices to show that themonolith of B/λ satisfies 1.

Assuming first that B is finite, let B = C/θ, where C ⊆ Am withm minimal. Let ψ be the unique cover of θ. By (i) it suffices to showthat ♯ψ/θ ≤ |A|. Let η ∈ ConC be the kernel of the projection tothe first coordinate, and η′ the kernel of the projection to the othercoordinates. Since η′ � θ by the minimality of m,

ψ/θց ψ ∧ η′/θ ∧ η′ ր η ∨ (ψ ∧ η′)/η ∨ (θ ∧ η′) ⊆ 1/η.

EXERCISES 103

If α is non-Abelian then by Theorem 10.1 B ∈ HS(A) and 1.clearly holds. Hence we assume that [ψ, ψ] ≤ θ and thus by the resultsof Chapter 6, θ and ψ ∧ η′ permute and the result follows from (ii).

Now assume that B is infinite. As above we may assume B is sub-directly irreducible with monolith α. Since B ∈ V (A), which is locallyfinite, Lemma 10.2 implies that B is a subalgebra of an ultraproductof finite algebras Bi ∈ HS(B). Lemma 10.2 also tells us that if µiis the monolith of Bi, then α is contained in the ultraproduct of theµi’s restricted to B. Since ♯µi/0 ≤ |A| for each i by the finite case.♯µ/0 ≤ |A|, from which 1. follows. The other parts of the theoremfollow easily. �

Exercises

1. Let C ⊆∏n

i=1Ai be a subdirect representation of C. Let α ≻β in Con C and let ηi be the projection congruences. Provethat there is an i such that ηi ≤ (β : α). Thus, in the situationof Theorem 10.5, if B ∈ V (A) is subdirectly irreducible, andA and B are finite, then there exists a nilpotent congruenceθ ∈ Con B such that θ =

∧ni=1(θ ∨ ηi).

CHAPTER 11

Joins and Products of Modular Varieties

The two main results of this chapter are easy but not obvious. Thefirst bounds the size of the subdirectly irreducible algebras in the join oftwo modular varieties. It was noticed in the case of finitely generatedvarieties in Freese-McKenzie [29], then it turned out to be true forrings (McKenzie [64]) so it gradually became obvious. An example isgiven showing that result is false without the modularity assumption.The second result shows that if two varieties intersect trivially and oneis solvable then they are independent. A theorem of C. Herrmann isderived as a corollary.

Theorem 11.1. Let V = V0 ∨ V1 be a modular variety. If Vi isresidually ≤ λi, i= 0, 1, then V is residually ≤ λ0λ1.

Proof. This theorem is an immediate consequence of the followinglemma: if V = V0∨V1 is modular and B ∈ V is subdirectly irreducible,then B = C/θ where C is a subdirect product of subdirectly irreduciblealgebras A0 and A1, Ai ∈ Vi.

To see the lemma let B ∈ V be subdirectly irreducible and assumeB 6∈ V0 ∪ V1. Then B = C/θ where C ⊆ A0 × A1, Ai ∈ Vi. Letηi ∈ Con(C) be the projection kernels. Using the upper continuity ofCon(C), choose θ0 and θ1 ∈ Con(C) maximal such that θi ≥ ηi andθ0∧ θ1 ≤ θ. The proof of the lemma will be done when we show θ0 andθ1 are completely meet irreducible.

First note that (θ ∨ θ0) ∧ (θ ∨ θ1) > θ. For otherwise B ∈ V0

or V1. Moreover, (θ ∨ θ0) ∧ (θ ∨ θ1)/θ ց θ1 ∧ (θ ∨ θ0)/θ ∧ θ1. Nowθ ∧ θ1 = θ0 ∧ θ1 since otherwise we could replace θ0 with θ0 ∨ (θ ∧ θ1)and violate the maximality of θ0. Hence θ1∧(θ∨θ0)/θ0∧θ1 is isomorphicto (θ ∨ θ0)∧ (θ ∨ θ1)/θ. Since θ is completely meet irreducible, there isa ϕ ≻ θ0 ∧ θ1 such that θ0 ∧ θ1 < α ≤ θ1∧ (θ∨ θ0) implies ϕ ≤ α. Sinceθ1 ∧ (θ ∨ θ0)/θ0 ∧ θ1 ր (θ0 ∨ θ1) ∧ (θ0 ∨ θ)/θ0, θ0 ≺ θ0 ∨ ϕ. Supposeψ ∈ Con(C) satisfies ψ ∧ (θ0 ∨ ϕ) = θ0. Then ψ � ϕ, which impliesψ ∧ θ1 ∧ (θ ∨ θ0) = θ0 ∧ θ1. Now

(θ ∨ θ0) ∧ ((ψ ∧ θ1) ∨ θ) = θ ∨ (ψ ∧ θ1 ∧ (θ ∨ θ0)) = θ.

105

106 11. JOINS AND PRODUCTS OF MODULAR VARIETIES

Since θ is meet irreducible, either θ ≥ θ0 ≥ η0, a contradiction, orψ ∧ θ1 ≤ θ. But then ψ = θ0 by the maximality of θ0. This proves θ0is completely meet irreducible. �

Example 11.2. Polin constructed a nonmodular variety V, whosecongruence lattices nevertheless did satisfy nontrivial lattice identities,solving an old problem. His variety is generated by a four elementalgebra A with a binary, two unary, and a nullary operation (see Polin[74], Day-Freese [23]). Now A = A0 ×A1 with each Ai equivalent tothe two element Boolean algebra. Hence V = V0 ∨ V1 with V = V (Ai)and Vi is distributive, permutable, and residually ≤ 2. Nevertheless Vis nonmodular and residually large. Thus modularity is necessary inTheorem 11.1. This example also shows that modularity is necessaryin the result of Hagemann and Herrmann (see Exercise 8.2), that thejoin of two distributive varieties in a modular variety is distributive.

Two varieties V0 and V1 of the same type are independent if there isa term t(x0, x1) such that Vi satisfies t(x0, x1) = xi, i = 0, 1. If V0,V1 ⊆V, we write V = V0⊗V1 provided that V0 and V1 are independent andV = V0 ∨ V1 (i.e. the variety generated by V0 ∪ V1 is V). Note thatif t is as above and if σi(x1, ..., xn), i = 0, 1, are terms for V, thenτ = t(σ0, σ1) is a term whose interpretation in Ai agrees with that ofσi if Ai ∈ Vi, i = 0, 1. From this it follows that V0 ⊗ V1 defined aboveis equivalent (in the technical sense; see Taylor [80] p. 354) to V0⊗V1

as defined by Taylor.The equation V = V0⊗V1 has strong structural implications for V:

see Gratzer-Lakser-Plonka [35] and Taylor [80], pp. 357–8. In partic-ular, each A ∈ V can be decomposed as A = A0 ×A1 with Ai ∈ Vi.Moreover Con(A) ∼= Con(A0)×Con(A1). Any (weak) Mal’cev con-dition which holds for both V0 and V1 will hold for V0 ⊗ V1.

Theorem 11.3. Let V be a modular variety. Let V = V0∨V1, whereV1 is solvable and V0∩V1 contains only the one-element algebras. ThenV = V0 ⊗ V1.

Proof. First we will show that if A ∈ V is a subdirect productof A0 and A1, with Ai ∈ Vi, then A = A0 × A1. To see this letηi ∈ Con(A) be the projection kernel, i = 0, 1. Then η0 ∧ η1 = 0and, since A/η0 ∨ η1 ∈ η0 ∩ η1, η0 ∨ η1 = 1. Since Ai is solvable,[1]k ≤ η1 in Con(A) for some k. ([1]k is defined in Definition 6.1.)Hence [η0]

k ≤ [1]k ∧ η0 ≤ η1 ∧ η0 = 0. Thus η0 is solvable and sopermutes with η1 by Theorem 6.2, showing that A = A0 ×A1.

Since V = V0 ∨ V1, FV(2) is a subdirect product of FV0(2) and

FV1(2). Thus FV(2) = FV0

(2) × FV1(2). Let x and y be the free

11. JOINS AND PRODUCTS OF MODULAR VARIETIES 107

generators of FV(2) and let ηi be the projection kernels. Since η0 ◦η1 =1, x η0 t(x, y) η1 y for some t(x, y) ∈ FV(2). Consequently V0 satisfiesx = t(x, y) and V1 satisfies y = t(x, y), proving the theorem. �

Since a distributive variety and a solvable variety can have only theone-element algebra in common, the last theorem implies that if D is adistributive variety and S is solvable variety and M = D∨ S then M =D⊗ S. Hence we have the following corollary, due to Herrmann [45].

Corollary 11.4. Let M = D∨A with M a modular variety, D adistributive variety and A an Abelian variety. Then M = D⊗A. �

We remark that for each of the congruence identities ε consideredin Chapter 8, if V = V0 ∨ V1 is modular, then V satisfies ε if and onlyboth V0 and V1 satisfy ε. This is due to the fact that “hereditary ε” isa property preserved under the formation of subdirect products of twofactors, and homomorphic images.

CHAPTER 12

Strictly Simple Algebras

We mean by a strictly simple algebra a finite simple algebra hav-ing more than one element, which is generated by any two of its dis-tinct elements. (This last condition is equivalent to saying that anyproper subalgebra has only one element.) A minimal variety is justan equationally complete variety, i.e., a variety possessing exactly twosubvarieties, including itself. It is obvious that each locally finite min-imal variety is generated by a strictly simple algebra. Surprisingly, theconverse of this statement is very nearly true in the domain of modularvarieties.

We shall also be concerned in this section with the spectrum setsand fine spectrum functions of varieties. We define

Spec(V) = {n < ω : for some A ∈ V, |A| = n}

nV(κ) = (the number of isomorphism types of algebras

in V of cardinality κ)

Notice that Spec(V) ⊆ ω, whereas nV is a function defined on all car-dinal numbers, finite and infinite.

The study of strictly simple algebras in modular varieties, and thespectra and fine spectra of modular varieties generated by strictly sim-ple algebras, produced some interesting results. We collect them hereand supply proofs which are relatively easy compared to those in theliterature.

The following was first proved for permutable varieties in McKenzie[62] (but Smith [79], Chapter 5, came very close to proving it), and formodular varieties later by C. Herrmann. The ternary discriminatoroperation on a set A is operation t : A3 → A given by t(a, b, c) = a ifa 6= b and c if a = b. A variety V is called a discriminator varietyif there is a term t(x, y, z) for V such that the corresponding termoperation is a ternary discriminator operation on each algebra in agenerating set of V. A finite algebra such that the ternary discriminatoroperation is a term operation is called quasiprimal.

Theorem 12.1. (1) Every strictly simple algebra in a modularvariety generates either an Abelian or a distributive subvariety.

109

110 12. STRICTLY SIMPLE ALGEBRAS

(2) IfA is strictly simple andV(A) is permutable and non-Abelian,then A is a quasiprimal algebra.

Proof. Suppose first that A is strictly simple, V = V (A) is mod-ular, and A is not affine. Obviously, A (and its subalgebras can haveonly one element) satisfies the congruence identity [x, y] = x · y (in factCon A = 0, 1 and [1, 1] = 1). Thus A and its subalgebras are neutral.Since A is finite, FV (A)(3) is a finite subdirectly power of A. Now, byExercise 8.2, FV (A)(3) is neutral, which implies its congruence latticeis distributive by Exercise 8.1. Thus, by Jonsson’s Theorem 2.1, V (A)is distributive.

For statement (2), suppose thatA is strictly simple and non-Abelianand V (A) is permutable. Then by (1) V (A) is arithmetic (distribu-tive and permutable), and by Jonsson’s theorem every subdirectly ir-reducible algebra in the variety is isomorphic to A, hence hereditarilysimple. So (2) follows from Pixley’s characterization of quasiprimalalgebras [72] (or see Burris-Sankappanavar [10] p.173). �

Corollary 12.2. Let F be a finite, simple, non-Abelian algebrain a modular variety and F′ = (F, c)c∈F its inessential expansion byconstants. Then we have

(1) V(F′) is distributive.(2) If V(F) is permutable, then the ternary discriminator opera-

tion is a polynomial operation of F.

The following amazing result of R. Quackenbush [76], (but he dis-covered it in 1976) stood for four years with a difficult proof until aneasy one was found in McKenzie [63]. Modularity is not implicitlyassumed. The proofs lie outside commutator theory.

Theorem 12.3. The following are equivalent for any finite algebraA, where |A| = n > 1.

(1) A is strictly simple and V(A) is permutable.(2) Spec(V(A)) = {nk : k < ω}.

Quackenbush has used his theorem in [77] to give a relatively easyproof of I. Rosenberg’s classification of pre-complete clones of opera-tions on a finite set.

If A is strictly simple and V (A) = V is distributive, then V isminimal by Jonsson’s Theorem. If V is also permutable, then Spec(V) isgiven by Theorem 12.3(ii), and nV(m) = 1 for finite m in the spectrum,since A is quasiprimal. On the other hand, if V is distributive but notpermutable, then most of the obvious questions concerning Spec(V)and the finite values of nV are completely open. [S. Burris proved in [7]

12. STRICTLY SIMPLE ALGEBRAS 111

that for infinite λ, any distributive variety containing a simple algebraof cardinality at most λ, has nV(λ) ≥ 2(λ) with equality, of course, ifV has at most λ basic operations.]

On the Abelian side of Theorem 12.1, there are some interestingfacts which we now proceed to derive under the rubric of theory devel-oped in Chapter 9. It is easily seen, first of all, that any finite simpleaffine algebra is strictly simple, the condition on subalgebras beingredundant. Let us recall the concept of the linearization, A∇, of anAbelian algebra A, from Proposition 9.18. The history and contribu-tors to the following theorem will be discussed after we prove it.

Theorem 12.4. Let Q be a finite, simple, Abelian algebra of nelements and let V = V (Q). Then n is a prime power. Furthermore

(1) If Q has an idempotent element, then(i) Every finite algebra in V is (isomorphic to) a power of Q

and every infinite algebra in V is a Boolean power of Q.Thus V is minimal.

(ii) All algebras in V of size λ are isomorphic, for each λ.(2) If Q has no idempotent element, then

(i) Every finite algebra in V is a power of Q or of Q∇ andevery algebra in V is a Boolean power of Q or of Q∇.V(Q∇) is the unique nontrivial proper subvariety of V.

(ii) nV(λ) = 2 if λ = nk (k < ω) and is 0 otherwise.

Proof. First we define the concept of isotopy. If A, B and C arealgebras in a variety, we say A is isotopic to B via C provided C×Ais isomorphic to C×B via an isomorphism which commutes with thefirst projection (i.e. (c, a) 7→ (c, a) for some b ∈ B). Notice that in amodular variety this is equivalent to

there is a β ∈ Con(C×A) with (C×A)/β ∼= B(∗)

and β a complement of η0.

(Here we are using Gumm’s result that every congruence of a directproduct permutes with the factor congruences, see Exercise 5.4). Noticealso that if A is isotopic to B via C and C has an idempotent element,then A ∼= B.

Now Q∇ = (Q×Q)/∆1,1 and since ∆1,1 and η0 are complements wehave thatQ andQ∇ are isotopic via Q. In particular Q×Q∇

∼= Q×Q,and if Q has an idempotent Q ∼= Qnabla.

Now suppose B ∈ V (Q) is finite and subdirectly irreducible. SinceQ has no proper subalgebras and generates a permutable variety, itsfinitely generated free algebras are direct powers of Q. Hence there is

112 12. STRICTLY SIMPLE ALGEBRAS

a k with Qk → B and we prove by induction on k that B is eitherQ or Q∇. Since Q is simple, k = 1 is easy. So assume k > 1. ThenQk ∼= Q×Qk−1. Let θ ∈ Con(Q∇×Q

k−1) be the kernel of the map ontoB. Since the coatoms of ConQk meet to 0, ConQk ∼= ConQ∇×Qk−1

is a complemented modular lattice and since θ is meet irreducible, thisimplies it is also a coatom. Let η′0 be the kernel of Q∇×Qk−1 → Qk−1

and define γ = (η0∧θ)∨η′0. By the minimality of k, θ 6≥ η′0 (and θ 6= η0)

and so γ is a coatom above η′0. By induction C = Q∇×Qk−1/γ is either

Q or Q∇. Also each pair from η0, θ, and γ is complementary in theinterval 1/η0 ∧ θ. By (*) we see that B is isotopic to C via Q∇. ButQ∇ has an idempotent. Hence B ∼= C, proving that Q and Q∇ are theonly subdirectly irreducible algebras in V (Q).

If A ∈ V (Q) is finite, then it is a finite subdirect product of Qand Q∇. The minimality of Q and Q∇ and permutability imply thatA ∈ P{Q,Q∇}. Since Q∇ ×Q ∼= Q×Q. A is a direct power of Q orof Q∇.

Let R = R(V (Q)) and M0 = M(1Q, 0) = M(Q, 0) where 0 isa fixed but arbitrary element of Q. Since Con Q = Con M0 byProposition 9.19(i), M0 is simple. It is a faithful R-module because Qgenerates V (Q). Thus R is a finite simple ring, and hence isomorphicto the ring of m by m matrices over a finite field GF(pr). Hencen = |Q| = |M0| = prm.

By Jacobson [52], Theorem 4.4, p.208, every R-module is isomor-

phic to a direct sum M(λ)0 for some cardinal λ.

If B ∈ V (Q) has an idempotent element e, then M = M(B, e) ∼=M0

(λ) and B = A(M, ϕ(e)) ∼= A(M0(λ), 0) by Theorem 9.16, so B is

determined up to isomorphism by its cardinality. Notice that if Q hasan idempotent element, then FV (Q)(1) does and hence every algebra inV (Q) does. This proves (1), since there are Boolean powers of everyinfinite cardinality.

To complete the proof we will show that if B ∈ V (Q) has infinite

cardinality λ and no idempotent element, then B ∼= A(M0(λ), 0)×Q.

Indeed we have B ∼= A(M0(λ), ψ) and we can assume the isomor-

phism is equality. The rang of ψ is a finite submodule of M0(λ),

hence contained in a finite sum of the factors. So we can write B =A(M0

(λ) × M(k), ψ), with k < ω and the range of ψ contained inthe second factor. Since the projection kernels are congruences of B,this implies B ∼= A(M0

(λ), 0) × A(M(k), ψ). The second factor is fi-nite without idempotent hence isomorphic to Qk by the above. SinceQk ∼= Qk−1 ×Q, we obtain B ∼= C ×Q where C has an idempotent.Hence C ∼= A(M0

(λ), 0), completing the proof. �

12. STRICTLY SIMPLE ALGEBRAS 113

Steven Givant in [31, 32] and E. A. Palyutin in [70, 71] describedin detail all varieties of algebras of countable type which are categoricalin some infinite power. The varieties V (Q∇) with Q strictly simpleAbelian are precisely the modular varieties in Givant’s classification.(The nonmodular ones can be constructed in a canonical way frompermutational representations of finite groups.) Note that V (Q) isessentially of finite type, whatever its actual type, since every term isequal in Q to one obtained by composition from binary terms and d,by Theorem 9.16.

The conclusions in Theorem 12.4 which concern the finite algebrasin V (Q) were proved in part by Smith [79], Chapter 5, and in fullby Clark and Krauss [12], [13] and Quackenbush [76]. Categoricityof V (Q∇) in all powers was then proved by Clark-Krauss [14]. Thefact that nV(λ) = 2 for all infinite λ when Q is not isomorphic to Q∇

follows from Corollary 2.8 and Lemma 2.9 of Clark-Krauss [14]. Ourproof of Theorem 12.4 has not appeared before.

CHAPTER 13

Mal’cev Conditions for Lattice Equations

Theorems 2.1(2) and 2.2 give Jonsson’s and Day’s conditions on avariety equivalent to the variety having distributive or modular congru-ences. Both of these conditions assert the existence of terms in a fixednumber of variables (3 for distributivity, 4 for modularity) which satisfycertain equations. Conditions of this form are known as Mal’cev con-ditions. In the light of Jonsson’s and Day’s results, it would be naturalto conjecture that every lattice identity has such a Mal’cev condition.A. Pixley and R. Wille were able to prove that each lattice has a it weakMal’cev condition. A it weak Mal’cev condition is the conjunction ofMal’cev conditions Cn, n ∈ ω, where Cn implies Cm if n ≥ m. SeeTaylor [80] for a precise discussion of these concepts. Several questionsremain open. it Does every lattice identity correspond to a Mal’cevcondition? Are there any nontrivial identities, other then distributiv-ity and modularity, which have Mal’cev conditions? Certain identitieswere shown to have Mal’cev conditions, for example, Gedeonova [30]and Mederly [66], but Day [20], following Nation, shows that theseidentities were in fact equivalent to modularity for the congruence lat-tice of varieties of varieties algebras. In this section we shall answer thesecond question above. We use the commutator to sketch a proof thatare infinitely many lattice identities each having a Mal’cev condition.Moreover, these identities are inequivalent in the strong since that forany two of them there is a modular variety whose congruence latticessatisfy one of these identities but not the other. In this chapter weassume the reader has some familiarity with Mal’cev conditions andthe theory of modular lattices.

A subset {ai : i = 1, . . . , n}∪{cij : 1 ≤ i, j ≤ n, i 6= j} of a modularlattice is called a (von Neumann) n–frame provided.

ai ∧∨

j 6=i

aj =

n∧

k=1

ak

(cij ∨ cjk) ∧ (ai ∨ ak) = cik

cij = cji

115

116 13. MAL’CEV CONDITIONS FOR LATTICE EQUATIONS

for distinct i, j and k, and ai, aj , cij generates a copy of M3 fori 6= j. These n–frames play an important role in the theory of modularlattices. They are projective, i.e.,the free modular lattice generated byan n–frame is a projective lattice. This fact was first proved by Huhn[48]; see also Freese [25]. Moreover, if L is a modular lattice containingan n–frame, n ≥ 4, we let R = {x ∈ L : x∨a2 = a1∨a2, x∧a2 = a1∧a2}and define an addition x⊕ y on R by

x⊕ y = [((x ∨ c13) ∧ (a2 ∨ a3)) ∨ ((y ∨ a3) ∧ (a2 ∨ c13))] ∧ (a1 ∨ a2)

Multiplication on R is defined by a similar formula. The details ofthese formulae are not important for our arguments here. Under theseoperations R is a ring with a1 as its zero and c12 as its unit.

The next lemma is the key to our result.

Lemma 13.1. If θ1, . . . , θn, θ12, θ13, . . . , θn−1,n is an n–frame in ConAin a modular variety, then any two congruences in the interval between∧

θi and∨

θi permute.

Proof. By additivity [∨

θi,∨

θi] =∨

i,j[θi, θj]. If i 6= j, then

[θi, θj ] ≤ θi ∧ θj =∧

θk. Since θi is part of an M3 with least ele-ment

∧

θk, [θi, θj] ≤∧

θk. Thus [∨

θi,∨

θi] ≤∧

θi and the lemma nowfollows easily from Theorem 6.2. �

Let x = (x1, . . . , xn, x12, . . . , xn,n−1) be a vector of lattice variables.The fact that n–frames are projective implies that there are latticeterms ai(x), cij(x), in these variables such that any interpretation ofthese terms into a modular lattice yields a (possibly degenerate) n–frame and if the variables xi and xij are substituted into a frame ai,cij the value of ai(x) is ai and that of cij(x) is cij. Let εp be the lat-tice equation which expresses the fact that the ring associated with theframe {ai(x), cij(x)} satisfies p · 1 = 0, where p is a prime. Here p · 1stands for the sum of p copies of 1. By the above formula for addition,this can be expressed as a lattice equation. If V is the variety of allvector spaces over a field F, then using elementary linear algebra onecan show that the ring associated with any n–frame in a congruencelattice (= subspace lattice) in V has the same characteristic as thatof F. Thus if p and q are distinct primes there is a modular variety(the variety of vector spaces over the field with p elements) which sat-isfies εp but fails εq. Hence the εp’s are pairwise inequivalent even forcongruences on varieties of algebras.

Theorem 13.2. For each prime p, the condition that the congru-ence lattice of a variety of algebras are modular and satisfy εp is defin-able by a Mal’cev condition.

13. MAL’CEV CONDITIONS FOR LATTICE EQUATIONS 117

Proof. In order to understand the proof we need to review theprocedure for obtaining the weak Mal’cev condition associated witha lattice inequality. This is best done by way of example. We willillustrate this with the distributive law, the simplest nontrivial latticeequation. The distributive law is equivalent to the following inequality.

(1) x ∧ (y ∨ z) ≤ (x ∧ y) ∨ (x ∧ z)

Recall that a weak Mal’cev condition consists of the conjunction ofstronger Mal’cev conditions Cn, n < ω. To illustrate the procedure forobtaining Cn, let V be a distributive variety and let S be a set (whosesize will be determinate later) of letters containing a and b, and letF = FV(S) be the free V algebra generated by S. In the first halfof the procedure we associate congruences of F with the variables of(1) in such a way that 〈a, b〉 is in the left side of (1). Since we want〈a, b〉 ∈ x∧ (y∨ z), we must have 〈a, b〉 ∈ x, and 〈a, b〉 ∈ y∨ z. In orderto insure that the latter occurs we include in S letters a1, . . . , an suchthat 〈a, a1〉 ∈ y, 〈ai, ai+1〉 ∈ y, if i is even 〈ai, ai+1〉 ∈ z, if i is odd, and〈an, b〉 is in y if n is even, in z if n is odd. (The n here is the sameas the subscript of Cn.) Thus in order to insure that 〈a, b〉 is in theleft side of (1) we take S = a, b, a1, . . . , an, and we take the followingsubstitutions for x, y and z.

x 7→ Cg (a, b)

y 7→∨

i even

Cg (ai, ai+1)(2)

z 7→∨

i odd

Cg (ai, ai+1)

Here we take a0 = a and an+1 = b. This is the first half of the procedurefor obtaining Cn.

The procedure for an arbitrary lattice inequality is similar. In thegeneral step if we have c, d ∈ S with 〈c, d〉 ∈ w, where w is a subterm ofthe left side of the inequality, then if w is a meet, say w = w1∧w2, thenwe put 〈c, d〉 in both w1 and w2. If w = w1 ∨w2 then we add c1, . . . , cnto S and put 〈ci, ci+1〉 into w1 or w2, depending on whether i is evenor odd. The procedure stops when we reach the lattice generators. Atthis point, associated with each lattice generator is a list of pairs ofelements of S. We then associate with this generator the congruenceon FV(S) generated by this list of pairs. It is easy to see that, underthis interpretation, 〈a, b〉 is in the left side of the inequality.


Now for the second half of the procedure. Let x, y and z denotethe value of these letters under the substitution (2), i.e., let x denoteCgF (a, b), etc. Thus 〈a, b〉 ∈ x∧(y∨z) and since we are assuming that Vsatisfies (1), we must have 〈a, b〉 ∈ (x∧y)∨(x∧z). Hence there must beelements t0(a, a1, . . . , an, b), . . . , tm(a, a1, . . . , an, b) of FV(S) such thatt0 = a, tm = b, 〈ti, ti+1〉 ∈ x ∧ y, if i is even, 〈ti, ti+1〉 ∈ x ∧ z if iis odd. Since x = Cg (a, b), for each i we have ti(a, a1, . . . , an, a) xti(a, a1, . . . , an, b) x ti+1(a, a1, . . . , an, b) x ti+1(a, a1, . . . , an, a). SinceCg (a, b) is the trivial relation on the subalgebra of F generated by{a, a1, . . . , an}, we have

(3) ti(a, a1, . . . , an, a) = ti+1(a, a1, . . . , an, a)

for i = 0, . . . , m−1. Similarly, assuming n is odd, we obtain for i even,

(4) ti(a, a, a2, a2, a4, a4, . . . , b) = ti+1(a, a, a2, a2, a4, a4, . . . , b)

and for i odd

(5) ti(a, a1, a1, a3, a3, . . . , b, b) = ti+1(a, a1, a1, a3, a3, . . . , b, b)

Now the Mal’cev condition Cn is the following assertion for a va-riety V: for some m < ω there are V-terms t0(a, a1, . . . , an, b),. . . ,tm(a, a1, . . . , an, b) such that t0 = a, tm = b and V satisfies (3), (4),and (5). The above arguments show that if V is a distributive varietythen V satisfies Cn for all n. It is not difficult to prove the converse: ifV satisfies Cn for all n, then V is distributive. Jonsson’s Mal’cev con-dition is C1. His theorem (Theorem 2.1) is the much stronger resultthat V is distributive if and only if it satisfies C1.

The same idea applies in the general case. Namely if we have alattice inequality

(6) v(x1, . . . , xk) ≤ u(x1, . . . , xk)

then following the first half of the procedure detailed above we obtain aset S, containing a and b, and a map, which we denote by τ , from eachlattice variable xi into a congruence on FV(S) generated by a partitionon S, such that 〈a, b〉 is in the left side of (6). If V satisfies (6) then〈a, b〉 is in the right side. Just as in the distributive case this impliesthat there are certain identities involving the variables S.

Thus V satisfies a lattice inequality, such as (6), if and only if Vsatisfies Cn for all n. An inequality such as (6) has a Mal’cev conditionif there is a fixed n0 such that V satisfies (6) if and only if it satisfiesCn0

. Since if V satisfies (6) it satisfies all Cn, to show that (6) has aMal’cev condition it suffices to show that there is a fixed n0 such thatif V fails (6) then V fails Cn0

.

13. MAL’CEV CONDITIONS FOR LATTICE EQUATIONS 119

Now let V be a variety failing the conjunction of modularity and εp.If V is nonmodular then the C2 of Day’s Mal’cev condition fails in V.Thus we may assume that V is modular and fails εp. We claim that theC1 associated with εp fails in V. Since εp fails there is an algebra A inV and congruences θi, θij which fail εp. By the remarks above, ai(θ),cij(θ) is a frame in Con(A) and if we substitute the variables of εp intothis frame, instead of into the θi, θij , the value of the two sides of εpremain the same. Thus, by changing notation, we may assume there isa substitution, xi 7→ θi, xij 7→ θij , of the variables into a frame {θi, θij}of Con(A) such that εp fails. Hence there are elements a, b ∈ A with〈a, b〉 in the left side of εp but not in the right. Now by Lemma 13.1all of the subterms of the terms of ε permute. Now we build a subsetT ⊆ Ain a manner analogous to the procedure for forming S in theweak Mal’cev condition C1 for εp described above. We start with a,b ∈ T . Let v be the left side of εp. Then 〈a, b〉 ∈ v. Now supposethat c, d ∈ T , and 〈c, d〉 ∈ w = w1 ∨ w2, where w is a subterm of v.Since w = w1 ◦ w2, there exists e ∈ A such that c w1 e w2 d. Weadd e to T . Since T is formed in a manner similar to S, there is asurjection σ : S → T . Moreover if y is one of the lattice variablesof v and if 〈c, d〉, (c, d ∈ S) is in the congruence of FV(S) associatedwith y (see the description of C1 above) then 〈σ(c), σ(d)〉 is in thecongruence on A associated with y (θi if y = xi, θij if y = xij). Thatis, the image under σ of the partition on S associated with y lies in thecongruence on A associated with y. If V satisfied C1 it would followfrom the construction of C1 that 〈a, b〉 would be in the right side of εp,a contradiction. Thus C1 fails in V. �

The lattice equations satisfied by the lattice of submodules (= con-gruence lattice) of a variety of modules were extensively studied inHutchinson-Czedli [51]. Using these ideas we can prove that for eachlattice equation ε there is another equation ε′, which is equivalent to εfor varieties of modules, and such that the conjunction of ε′ and mod-ularity is definable by a Mal’cev condition.

A recent paper of Day and Kiss defines a “canonical” n–frame inFV(n + 1) when V is a modular variety. They show that the ringassociated with this frame equals R(V). Thus the equations εp aresignificant for V in that V satisfies εp if and only if r ·1 = 0 in R(V), forany p (not necessarily a prime). Thus our Mal’cev condition are closelyconnected to the characteristic of R(V). Some of these connections aredeveloped in the exercises.


Exercises

1. (Jonsson, unpublished) Let V be a modular variety, A ∈ V andαi, γij ∈ Con(A) be a n–frame, n ≥ 3, such that α1 ∧α2 = 0.Let β, δ be in the ring associated with this frame (i.e., β∨α2 =α1∨α2 and β∧α2 = α1∧α2 and the same for δ) and define anaddition and multiplication + and × on the ring as follows:

β+δ = {〈x, y〉 ∈ A2 : ∃ z, t x β z α2 y, x α2 t δ y, z α1 T}

β × δ = {〈x, y〉 ∈ A2 : ∃ z, t x β z α2 y, x α1 t δ y, z γ12 t}.

Prove that the addition given here is the same as the additiondefined just above Lemma 13.1. (This exercise can be used toconstruct a fairly simple Mal’cev condition for εp.)

2. As mentioned at the end of the chapter, all the congruence lat-tice in V satisfy εn if and only if R(V) (and so also R(V, λ, κ))satisfies n · 1 = 0. Prove this for n = 3. You may use thatthe satisfaction of εn is equivalent to the fact that the ring ofevery frame satisfies n · 1 = 0.

CHAPTER 14

A Finite Basis Result

In this chapter we presents a generalization of Michael Vaughan-Lee’s finite basis [81]. The result is that a finite nilpotent algebra offinite type which is the direct product of algebras of prime power ordershas a finite basis for its identities. Vaughan-Lee proved this under theadditional assumption that the algebra had an equationally definedconstant.

Lemmas 7.3 through Corollary 7.7 prove some elementary factsabout nilpotent algebras. The reader may want to quickly review thembefore proceeding. Recall from Chapter 6 that (1, 1]0 = 1 and that(1, 1]k+1 = [1, (1, 1]k]. As in Chapter 7, we write (1]k for (1, 1]k. A isnilpotent if (1]k = 0 for some k. For a (modular) variety V we let Vnbe the subvariety consisting of all algebras in V satisfying (1]n = 0.

We need to show that Vn is finitely based relative to V. The nexttheorems prove this provided FV(2) is finite. We are interested in thecase V = V (A), A finite, where this condition is of course true.

Theorem 14.1. a ζA b if and only if

(1) f(d(r1(a, b), r1(b, b), c1), . . . , d(rn(a, b), rn(b, b), cn)) =

d(f(r1(a, b), . . . , rn(a, b)), f(r1(b, b), . . . , rn(b, b)), f(c))

and

(2) d(r(a, b), r(b, b), r(b, b)) = r(a, b)

for all basic operations f , all c = 〈c1, . . . , cn〉 ∈ An, and all binaryterm operations r and ri.

Proof. If a ζ b then (1) and (2) hold by Proposition 5.7. Nowsuppose that (1) and (2) hold. Then (1) holds whenever f is a termoperation. (1) and (2) imply

(3) d(d(r(a, b), r(b, b), c), c, e) = d(r(a, b), r(b, b), e)

121

122 14. A FINITE BASIS RESULT

for all c, e ∈ A, and r a binary term operation. In fact,

d(d(r(a, b), r(b, b), c, e)

= d(d(r(a, b), r(b, b), c), d(r(b, b), r(b, b), c), d(r(b, b), r(b, b), e)

= d(d(r(a, b), r(b, b), r(b, b)), r(b, b), e)

= d(r(a, b), r(b, b), e).

Now define a relation θ on A by

(4) u θ v ↔ u = d(r(a, b), r(b, b), v)

for some binary term operation r.Clearly v θ v (take r(x, y) = y). Suppose u θ v so that u =

d(r(a, b), r(b, b), v). Let s(a, b) = d(r(b, b), r(a, b), r(b, b)). Note thats(b, b) = r(b, b). Now using (1) backwards and then (2),

d(s(a, b), s(b, b), u)

= d(d(r(b, b), r(a, b), r(b, b)), d(r(b, b), r(b, b), r(b, b)), d(r(a, b), r(b, b), v))

= d(d(r(b, b), r(b, b), r(a, b)), d(r(a, b), r(b, b), r(b, b)), d(r(b, b), r(b, b), v))

= d(r(a, b), r(a, b), v)

= v.

Thus v θ u.Suppose u θ v θ w. Then u = d(r(a, b), r(b, b), v) and v = d(s(a, b), s(b, b), w).

Let t(a, b) = d(r(a, b), r(b, b), s(a, b)). Note t(b, b) = s(b, b) and

d(t(a, b), t(b, b), w)

= d(d(r(a, b), r(b, b), s(a, b)), d(r(b, b), r(b, b), s(b, b)), d(r(b, b), r(b, b), w))

= d(d(r(a, b), r(b, b), r(b, b)), r(b, b), d(s(a, b), s(b, b), w))

= d(r(a, b), r(b, b), v)

= u.

If ui θ vi, i = 1, . . . , n and f is an n-ary basic operation then (1)implies f(u1, . . . , un) θ f(v1, . . . , vn). Thus θ is a congruence. Since (2)implies d(a, b, b) = a (take r(x, y) = x), a θ b. Hence, Cg (a, b) ≤ θ.But if u θ v then u = d(r(a, b), r(b, b), v) Cg (a, b) d(r(b, b), r(b, b), v) =v. Thus θ = Cg (a, b).

14. A FINITE BASIS RESULT 123

Now if 〈ui, vi〉 ∈ Cg (a, b), then ui = d(ri(a, b), ri(b, b), vi). Hence, iff is an n–ary term operation and c = 〈c1, . . . , cn〉 ∈ A

n, then using (3)

f(d(u1, v1, c1), . . . , d(un, vn, cn))

= f(d(d(r1(a, b), r1(b, b), v1), v1, c1), . . . , d(d(rn(a, b), rn(b, b), vn), vn, cn))

= f(d(r1(a, b), r1(b, b), c1), . . . , d(rn(a, b), rn(b, b), cn))

= d(f(r1(a, b), . . . , rn(a, b)), f(r1(b, b), . . . , rn(b, b)), f(c)).

Now using (3) again with r(a, b) = f(r1(a, b), . . . , rn(a, b))

d(f(u1, . . . , un), f(v1, . . . , vn), f(c1, . . . , cn))

= d(f(d(r1(a, b), r1(b, b), v1), . . . , d(rn(a, b), rn(b, b), vn)), f(v), f(c))

= d(d(f(r1(a, b), . . . , rn(a, b)), f(r1(b, b), . . . , rn(b, b)), f(v)), f(v), f(c))

= d(f(r1(a, b), . . . , rn(a, b)), f(r1(b, b), . . . , rn(b, b)), f(c))

Hence, f(d(u1, v1, c1), . . . , d(un, vn, cn) = d(f(u), f(v), f(c)) for all cand all (u, v) ∈ Cg (a, b). Thus, by Proposition 5.7, [Cg (a, b), 1] = 0,and a ζA b, as claimed. �

As above let V be a modular variety and Vn = {A ∈ V : (1]n = 0 in A}.Define sets En of equations as follows. Let E0 = {x ≈ y} and let En+1

be the set of all equations of the form

f(d(r1(s, t), r1(t, t), z1), . . . , d(rk(s, t), rk(t, t), zk))

≈ d(f(r1(s, t), . . . , rk(s, t)), f(r1(t, t), . . . , rk(t, t)), f(z))

union the set of all equations of the form

d(r(s, t), r(t, t), r(t, t)) ≈ r(s, t)

where f is a basic operation, r and ri are binary terms, and s ≈ t ∈ En,and the zi are variables.

Theorem 14.2. Let A ∈ V. Then A satisfies En if and only ifA ∈ Vn.

Proof. Suppose that A ∈ V. Then A ∈ Vn+1 if and only ifA/ζ ∈ Vn. By induction this holds if and only if A/ζ satisfies En. Thisholds if and only if for all evaluations a and b of s ≈ t in En, a ζ b. Bythe last theorem, this holds if and only if A satisfies En+1. �

When FV(2) is finite, then there is a finite set S of binary terms[namely a set of representative terms for the elements of FV(2)] suchthat any binary term r(x, y) for V is equivalent to a term in S, i.e.,r(x, y) ≈ s(x, y) holds in V for some s(x, y) in S. In this case, En can


be defined using only the binary terms of S, and so can be taken to befinite. Thus we have the following corollary.

Corollary 14.3. In a modular variety V of finite type, if FV(2) isfinite, then there is a finite set of laws which, together with the identitiesof V, define Vn.

Lemma 14.4. Let A ∈ V and let 0 be arbitrary in A and S ⊆ Awith s ζA 0 for all s ∈ S. Suppose 〈a, 0〉 ∈

∨

S Cg (s, 0). Then

a =

m∑

k=1

rk(sk, 0)

for some m ∈ ω, sk ∈ S and rk(x, z) binary idempotent terms for V.

Proof. Let θ =∨

S Cg (s, 0). Then θ ≤ ζ and hence (0 : θ) = 1.By Theorem 9.9 (with β = θ and γ = 1) the interval θ/0 of Con A isisomorphic to the lattice of submodules of the R(V, 1)–moduleM(θ, 0).Let R = R(V, 1) = R(V). By Exercise 5 of Chapter 9 the image ofCg (s, 0) under this isomorphism is the submodule Rs. Thus M(θ, 0) =∨

s∈S Rs =∑

s∈S Rs. Since the elements of R = R(V) can be repre-sented by binary idempotent terms the result follows. �

By Theorem 6.2 a nilpotent variety is congruence permutable. Hencewe assume that A ∈ U where U is a permutable variety with Mal’cevterm d(x, y, z). Consider the free U algebra F generated by X ∪ z.Define u + v = d(u, z, v), u, v ∈ F. For x ∈ X let δx ∈ End(F)be such that xδx = z, yδx = y, y ∈ X − x, zδx = z. An ele-ment w = w(x1, . . . , xn, z) of F is a commutator if there is a setx1, . . . , xn ⊆ X such that w is in the subalgebra generated by this setand z and wδ = z, i = 1, . . . , n. A term corresponding to a commutatoris a called a commutator word.

Lemma 14.5. (Higman’s Lemma) Assume as above that F is thefree algebra for U on X ∪ z and in addition that U is nilpotent. If u,v ∈ F then there is a finite set C of commutators such that

(1) the identity u ≈ v, together with the identities of U, impliesthe law w(x, z) ≈ z for all w ∈ C,

(2) 〈u, v〉 is contained in the congruence generated by {〈w(x, z), z〉 :w ∈ C}.

Proof. For w ∈ F define wδ∗x = d(w,wδx, z). If S ⊆ X , δS isthe endomorphism of F with xδS = z, if x ∈ S, and fixing the othergenerators. Assume that X is linearly ordered. If S ⊆ X is finite, δ∗Sis the composition of the δS

∗’s, x ∈ S in the order of X .


Define

C = {d(u, v, z)δKδ∗L : K ∩ L = ∅, K ∪ L = X , L finite}

It is not hard to see and is in fact shown below that C is finite. IfS, T ⊆ X then, δSδT = δT δS, δSδ

∗T = δ∗T δS, and, if S ∩ T 6= ∅, then

wδ∗T δS = z. Now if K ∩ L = ∅ and L is finite then d(u, v, z)δKδ∗L is in

the subalgebra generated by L ∪ z and if x ∈ L, d(u, v, z)δKδ∗LδX = z

by the above. Thus C consists of commutators. Clearly the law u ≈ vimplies the law d(u, v, z) ≈ z, and hence the law d(u, v, z)δKδ

∗L ≈ z.

Let c = d(u, v, z), and suppose that c lies in the subalgebra gen-erated by S = {x1, . . . , xn, z}. Then cδ∗L = z unless L ⊆ S, andcδKδ

∗L = cδK1

δ∗L ifK∩S = K1∩S. Hence C = {cδKδ∗L : K∪L = S}. Let

θ =∨

w∈C Cg (w, z). We claim that (c, z) ∈ θ. We show by inductionon r that (cδKδ

∗L, z) ∈ θ when K ∩L = ∅, and K ∪L = {x1, . . . , xn−r}.

The case r = 0 is true by the definition of θ, and r = n is theclaim. Suppose the result is true for r. Let K ∩ L = {x1, . . . , xn−r−1},K∪L = {x1, . . . , xn−r−1}, K ∩L = ∅. LetM = K ∪{xn−r}. By induc-tion, (cδMδ

∗L, z) and (cδKδ

∗N , z) are in θ. Now cδMδ

∗L = cδKδ

∗Lδxn−r

andcδKδ

∗N = cδKδ

∗Lδ

∗xn−r

= d(cδKδ∗L, cδKδ

∗Lδxn−r

, z). Hence both of theseelements are θ-related to z. So

z θ d(cδKδ∗L, cδKδ

∗Lδxn−r

, z)

θ d(cδKδ∗L, cδMδ

∗L, z)

θ d(cδKδ∗L, z, z)

= cδKδ∗L

proving the claim. Now since we are in a nilpotent variety, we can useLemma 7.6 to see that (u, v) ∈ θ, proving the lemma. �

Lemma 14.6. Let U be a permutable variety and w(x1, . . . , xk, z) ∈FUn

(X ∪ z) = F. Then

w(x, z) = w(z, . . . , z) + c1 + · · ·+ cm

where each ci is a commutator (for U) and the sum is associated leftto right.

Proof. Induct on n. The result is clear for n = 0. So we assumethat (1]n+1 = 0 but (1] 6= 0 in Con F, and that the result is true forF/(1]n. Hence by Corollary 7.4.

w(x, z) = w(z, . . . , z) + c1 + · · ·+ cm + e(x, z)

where the ci’s are commutators for U and e(1]nz. Setting each xi to z wesee that w(z, . . . , z) = w(z, . . . , z) + e(z, . . . , z). Thus e(z, . . . , z) = z.


Suppose that e lies in the subalgebra generated by {x1, . . . , xr, z} anddefine

u(x, z) =∑

S⊆{x1,...,xr}

(−1)|S|eδS.

Now a congruence α is said to be fully invariant if αδ ⊆ α for allendomorphisms δ. By Proposition 4.4(1) the commutator of two fullyinvariant congruences is again fully invariant. From this it follows that(1]n is fully invariant. Thus since e (1]n z, eδS (1]n z for each S. ThuseδS ζ z and hence the order of the sum above is irrelevant. Clearlyu(x, z) is a commutator (one needs e(z) = z for the case r = 0). Now

e = u−∑

∅6=S⊆{x1,...,xr}

(−1)|S|eδS

By induction on r, each eδS, S 6= ∅, is a sum of commutators which liein z/ζ . Hence e = e1+ · · ·+et, with ei a commutator and ei ζ z. Hencew(x, z) = (w(z, . . . , z)+ c1+ · · ·+ cm)+(e1+ · · ·+et). But since ei ζ z,w(x, z) = w(z, . . . , z) + c1 + · · ·+ cm + e1 + · · ·+ et, as desired. �

Let U be a permutable variety and let A ∈ U be a nilpotent algebra.Let 0 ∈ A be fixed but arbitrary. For a ∈ A, we let ρa : A →A be defined by xρa = x + a, where + is with respect to 0. ByCorollary 7.4, ρa is a permutation of A. Let R(A) be the subgroup ofthe full symmetric group, Sym(A), generated by {ρa : a ∈ A}.

Lemma 14.7. If A is a finite nilpotent algebra then the order of ρadivides |A|.

Proof. Induct on the nilpotency class of A. The initial case istrivial. Let c ∈ A, and let n = |A/ζ |, m = |0/ζ |. Then |A| = nm byCorollary 7.5. By the inductive hypothesis, cρna ζ c. Hence, cρ

na = c+d,

for some d ∈ 0/ζ . Since d ∈ 0/ζ , (b+d)ρa = (b+d)+a = (b+a)+d =bρa + d. Thus cρ2na = (c + d)ρna = c + d + d. Repeating this argumentwe obtain cρnma = c+md = c, proving the lemma. �

Lemma 14.8. If A is a finite nilpotent algebra then R(A) is asolvable group and if A has prime power order then so does R(A).

Proof. We induct on the class of A. The lemma is clear for thetrivial algebra. As was proved in the last lemma, if c ∈ 0/ζ then(x + c)ρa = xρa + c. Hence for any ρ ∈ R(A), (x + c)ρ = xρ + c.Thus if x′ ζ x, then by Corollary 7.4 x′ = x+ c for some c ∈ 0/ζ , andx′ρ = xρ + c ζ xρ. It follows that there is a natural homomorphismfrom R(A) onto R(A/ζ) which maps ρa to ρa/ζ . If ρ is in the kernelK of this homomorphism then for all x there is a c ∈ 0/ζ such that


xρ = x + c. Hence if ρ1 and ρ2 are in K and xρi = x + ci, thenxρ1ρ2 = x+ c1 + c2 = x+ c2 + c1 = xρ2ρ1, i.e., K is an Abelian normalsubgroup. If ρ ∈ K then the order of ρ divides |0/ζ | since if xρ = x+ cthen xρk = x + kc. By induction R(A/ζ) is solvable and so R(A) issolvable. If A has prime power order then by induction R(A/ζ) hasprime power order and so does K. Hence R(A) has prime power orderand so is nilpotent. �

Theorem 14.9. Let V be a modular variety of finite type. If A ∈ V

is finite of prime power order and nilpotent then A is finitely based. Inaddition there is an integer M such that if w(x, z) is a commutator inmore then M variables then A satisfies w(x, z) ≈ z.

Proof. By Theorem 2.1(1) there is a finitely based permutablevariety U containing A. We let d(x, y, z) be a Mal’cev term for U willdenote a finitely based variety containing A. As the proof proceeds wewill add identities to the axioms of U in such a way that U is alwaysfinitely based and A ∈ U.

We proceed by induction on the block size of [1, 1]. If this is 1then A is a finite Abelian algebra of finite type. By Proposition 5.7 Asatisfies the laws

f(d(x1, y1, z1), . . . , d(xn, yn, zn)) ≈ d(f(x), f(y), f(z))

for f a basic operation. These laws imply the corresponding laws forf an arbitrary term. In particular, d commutes with itself and soby Lemma 5.6 defines a ternary Abelian group, and any term can bewritten as a sum (using d) of one and two variable terms, see equation(8) of chapter 9. It follows that the above set of laws of A togetherwith a finite basis for the two variable laws of A, forms a finite basisfor the laws of A. We leave the details and the proof that the laststatement of the theorem is true in this case with M = 3 to the reader.

Now suppose that (1]c = 0 but (1]c−1 6= 0. Of course there arefinitely many identities of A which imply that FV (A)(2) is finite. Weassume that the laws of U include these laws. By Corollary 14.3 thereis a finite set of laws which define Uc relative to U. We assume thatthe laws of U also include these laws so that U = Uc.

Let θ ∈ Con A with 0 ≺ θ ≤ (1]c−1 ≤ ζ . Recall from chapter4 that A(θ) is θ viewed as a subalgebra of A × A and that ∆θ,1 isthe congruence on A(θ) generated by {〈〈a, a〉, 〈b, b〉〉 : a, b ∈ A}. LetC = A(θ)/∆θ,1 and B = C×A/θ.

Lemma 14.10. C is Abelian. Moreover the block size of [1B, 1B] issmaller than that of [1A, 1A].


Proof. Let η0 and η1 be the projection congruences onA(θ). Theneasy calculations show that η0 +∆ = η1 +∆ = 1 for ∆ = ∆θ,1. Hencein Con A(θ), [1, 1] = [η0 + ∆, η1 + ∆] ≤ ∆ + η0η1 = ∆. Thus C isAbelian.

By Proposition 4.5, [1B, 1B] = [1C × 1A/θ, 1C × 1A/θ] = [1C, 1C] ×[1A/θ, 1A/θ] = 0C × [1A/θ, 1A/θ]. Hence the block size of [1B, 1B] equalsthe block size of [1A/θ, 1A/θ]. Since θ ≤ [1A, 1A] the block size of[1A/θ, 1A/θ] is the block size of [1A, 1A] divided by the block size of θ. �

Now by induction and Higman’s Lemma, B has, relative to thelaws of U, a finite basis w1(x, z) ≈ z, . . . , wm(x, z) ≈ z, where the wi’sare commutators. Moreover there is a K ∈ ω such that if w(x, z) is acommutator in more then K variables then B satisfies w(x, z) ≈ z.

Lemma 14.11. Let V be a modular variety and let F = FV(X) bethe free algebra generated by an infinite set X. Then the center of Fis a fully invariant congruence.

Proof. Let d be a Gumm difference term for V and let a, b ∈ Fwith a ζF b. Then by Theorem 14.1, a and b satisfy (1) and (2) ofTheorem 14.1 for any choice of the ci’s. There is a finite subset X0

of X such that a and b are in the subalgebra generated by X0. Letx1, . . . , xn be in X−X0. Then (1) holds with ci = xi, i = 1, . . . , n. Letδ be an endomorphism of F and let c1, . . . , cn ∈ F be arbitrary. Let δ′

be the endomorphism of F that agrees with δ on X − x1, . . . , xn andxiδ

′ = ci. Of course, aδ = aδ′ and bδ = bδ′. Now we have that (1) ofTheorem 14.1 holds with a and b replaced by aδ and bδ. It now followsfrom that theorem that aδ ζF bδ, as desired. �

Continuing with Theorem 14.9, we let F = FU(X ∪ z) where X isa countably infinite set. Recall that 0 is a fixed but arbitrary elementof A. Since B satisfies wi(x, z) ≈ z and A/θ ∈ V (B), for any substi-tution of the variables into A with z 7→ 0 we have wi(a, 0)θ0. Sinceθ is Abelian and minimal it follows from Exercise 3 of Chapter 9 orfrom Lemma 14.4 that the group associated with each block of θ is anelementary Abelian p-group. Since θ ≤ ζA we have, by Theorem 14.1,that there is a finite set of laws of A which imply that wi(x, z) ζF z,i = 1, . . . , m and that wi(x, z) has order p in F. We may assume U

satisfies these laws.Of course the laws of B are the elements in the fully invariant

congruence generated by the pairs 〈wi(x, z), z〉. By Lemma 7.6 this isthe congruence(5)∨

(Cg(wi(u1, . . . , un, u0), u0) : ui ∈ F ) =∨

Cg(d(wi(u, u0), u0, z), z)


Let wi(x1, . . . , xn, x0, z) = d(wi(x, x0), x0, z). Since wi(x, z) ζF z, wi(x, x0) ζFx0 by Lemma 14.11. Thus we have that wi(x, x0, z) ζF z.

Now since B satisfies wi(x, z) ≈ z and wi(x, z) is a commutator wehave

(6) b satisfies wi(x, x0, z) ≈ z

(7) U satisfies wi(z, z, z) ≈ z

Lemma 14.12. Suppose that B satisfies u(x, z) ≈ z. Then there areidempotent terms rj(x, z) such that

(8) u(x, z) =∑

rj(vj, z)

where each vj has the form wi(s1, . . . , sk, s0, z) for some s0, . . . , sk ∈ F .Moreover, rj(vj, z) ζF z.

Proof. Since u(x, z) ≈ z holds in B, < u(x, z), z > is in thecongruence given by (5), i.e., the congruence generated by the pairsalgorithmwi(s1, . . . , sn, s0, Z), z〉 with si ∈ F. Since wi(x, x0, z) ζF zand the center is a fully invariant congruence of F, the first part of theresult now follows from lemma 14.4. Since vi ζF z and rj is idempotentthe last statement is also true. �

Lemma 14.13. Suppose that B satisfies u(x1, . . . , xn, z) ≈ z andti(x, z) ≈ z, i = 1, . . . , n. If A satisfies u(z, . . . , z) ≈ z then it satisfies

u(t1(x, z), . . . , tn(x, z), z) ≈ z.

Proof. Let f be a homomorphism from F to A and let f(xi) = ai,i = 1, . . . , n, and let 0 ∈ A denote f(z). Since A/θ is in V (B) and Bsatisfies ti(x, z) ≈ z, ti(a, 0)θ0. Let ci = ti(a, 0), i = 1, . . . , n, so thatci θ 0. Then, since C = A(θ)/∆θ,1 satisfies u(x, z) ≈ z, we have inA(θ)

〈u(c1, . . . , cn, 0), 0〉 = 〈u(c1, . . . , cn, 0), u(0, . . . , 0)〉

= u(〈c1, 0〉, . . . , 〈cn, 0〉, 〈0, 0〉)

≡ 〈0, 0〉 mod ∆θ,1

By Theorem 4.9 this implies that 〈u(c1, . . . , cn, 0), 0〉 ∈ [θ, 1] = 0.Hence u(c1, . . . , cn, 0) = 0 in A, proving the lemma. �

Lemma 14.14. There is a finite set of laws of A which imply

u(a1 + v(x, z), a2, . . . , an, z) = u(a1, . . . , an, z)

for all ai ∈ F and for all u(x, z) and v(x, z) such that B satisfiesu(x, z) ≈ z, v(x, z) ≈ z, and U satisfies u(z, z) ≈ z.


Proof. Assume that B satisfies u(x, z) ≈ z, v(x, z) ≈ z, and thatU satisfies u(z) ≈ z. Since 〈v, z〉 is in the fully invariant congruencegenerated by the 〈wi, z〉, which are contained in ζF by Lemma 14.11,v ζF z. Thus by Proposition 5.7 and the fact u(z, z) = z we have

u(a1 + v, a2, . . . , an, z)

= u(d(v, z, a1), d(z, z, a2), . . . , d(z, z, an), d(z, z, z))

= d(u(v, z, . . . , z), u(z, . . . , z), u(a1, . . . , an, z))

= d(u(v, z, . . . , z), z, u(a1, . . . , an, z))

= u(a1, . . . , an, z) + u(v, z, . . . , z, z)

for all ai ∈ F. By Lemma 14.13, u(v, z, . . . , z) ≈ z is an identity ofA. Hence it suffices to show there is a finite set of laws of A whichimply all laws of the form u(v, z, . . . , z) ≈ z, where u and v satisfy thehypotheses of the lemma.

Since B satisfies u(x, z) ≈ z, we can use Lemma 14.12 to writeu(x, z) in the form (8) as described in the lemma. Writing this out infull gives

(9) u(x, z) =∑

rj(wij(sj1(x, z), . . . , s

jn(x, z), s

j0(x, z), z), z)

where the rj are idempotent. If we apply the endomorphism whichsends x1 to v and the other generators to z, we have

u(v, z, . . . , z) =∑

j

(wij (sj1(v, z, . . . , z), . . . , s

jn(v, z, . . . , z), s

j0(v, z, . . . , z), z), z)

Since v ζF z, sjk(v, z) ζF sjk(z, z). Now in general by Proposition 5.7if a ζ b then d(d(a, b, c), c, e) = d(a, b, e) (see the proof of (3) above).Thus using Proposition 5.7 we obtain the following where, of course,a− b = d(a, b, z) and a+ b = d(a, z, b)

(10)

u(v, z, . . . , z) =∑

rj(wij(sj1(v, z)−s

j1(z, z), . . . , s

j0(v, z)−s

j0(z, z), z), z)

+∑

rj(wij(sj1(z, z), . . . , s

j0(z, z), z), z)

Applying to (9) the endomorphism which sends all the generatorsto z we obtain

z = u(z, z) =∑

rj(wij(sj1(z, z), . . . , s

j0(z, z), z), z)


From this and (10) we see that it suffices to show that there is a finiteset of laws of A which imply the laws

(11) wi(ti(v, z), . . . , tn(v, z), t0(v, z), z) ≈ z

where tj(x1, . . . , xm) are idempotent.Since v ζF z, tj(v, z) ζF tj(z, z) = z. Thus by equation (3) or (7)

of chapter 9 (or by using Proposition 5.7) we have that the left side of(11) is the sum of elements of the form

(12) wi(z, . . . , z, tj(v, z), z, . . . , z).

Since B satisfies v(x, z) ≈ z, we have by Lemma 14.12 that v(x, z) =∑

rj(qj , z) where rj is idempotent and qj has the form wij (p1, . . . , pn, p0, z),pi ∈ F. Since rj(qj, z) ζF z by Lemma 14.12, (12) can be expressed asa sum of elements of the form

wi(z, . . . , z,mth

tm(rj(qj , z), z, . . . , z), . . . , z)

It follows that (11) is a consequence of the laws

(13) wi(z, . . . ,mth

tm(rj(wij(x, y, z), z), . . . , z), . . . , z) ≈ z

Since both tm and rj are idempotent and B satisfies wi(x, y, z) ≈ zby (6), B satisfies

tm(rj(wij (x, y, z), z), z, . . . , z) ≈ z

Thus by Lemma 14.13 equation (13) is a law of A.Since tm(rj(x, z), z, . . . , z) is binary and there are only a fixed num-

ber of wi, there are only finitely many laws of the form (13). Thiscompletes the proof of the lemma. �

Theorem 14.15. There is an N and a finite set of laws of A suchthat if these laws are added to the axioms of U then bj ∈ F implieswi(b1, . . . , bn, b0, z) is a sum of elements of the form ±wi(a1, . . . , an, a0, z)where each aj is a sum of at most N commutators and an element inthe subalgebra generated by z.

Proof. Recall that by Lemma 14.8, R(A) is a finite p-group. LetZp(R(A)) be the group ring of R(A) over the integers modulo p, i.e.,Zp(R(A)) =

∑

kρρ : kρ ∈ Zp with the multiplication induced fromR(A). The Jacobson radical of this ring is the augmentation ideal:∑

kρρ :∑

kρ = 0. For a proof of this fact see Satz 5.16 on page 484 ofHuppert [50]. Hence ρ−1 is in the radical. Since the ring is finite, theradical is nilpotent, i.e., there is an N such that the product of any Nelements of the radical is 0. Thus if ρ1, . . . , ρN ∈ R(A) then

(14) (ρ1 − 1)(ρ2 − 1) . . . (ρN − 1) = 0


in Zp(R(A)). Fix a, a1, . . . , an ∈ A. For S = {i < j < · · · < k}a subset of {1, .., N} let aS = ai + aj + · · · + ak associated left toright. We claim that the number of even subsets S with a = aS equalsmodulo p the number of odd subsets with a = aS. To see this let V bea vector space over Zp with the set A as a basis (be careful to note that0A 6= 0V). Let Zp(R(A)) act on V by extending the natural action ofR(A) on A. In (14) let ρi = ρai then expanding out (14) we have

∑

S

(−1)|S|ρaiρajρak = 0

in Zp(R(A)). Applying this to the vector space element 0A (which,recall, is a basis element of V) we obtain

(15) 0V =∑

S

(−1)|S|0AρS,

where 0AρS = 0Aρaiρaj . . . ρak = 0+ai+aj+· · ·+ak = ai+aj+· · ·+ak.Looking at the coefficient of a in (15) gives the claim.

Hence if w(x, z) ζF z and has order p then for all b2, . . . , bn ∈ A∑

S

(−1)|S|w(aS, b2, . . . , bn, 0) = 0

Thus A satisfies the finite set of laws of the form∑

S

(−1)|S|wi(x1, . . . , xj−1, yS, xj+1, . . . , xn, x0, z) ≈ z

where yS = yi + yj + · · ·+ yk. One term in this sum has y1 + · · ·+ yNin the jth place and this law allows us to express this term as a sum ofterms in which the jth entry has fewer than N summands. The lemmanow follows from Lemma 14.6. �

Now we are ready to complete the proof of Theorem 14.9. We as-sume that the finite set of laws which imply the conclusions of Lemma 14.14and Theorem 14.15 is contained in the laws of U. Let p be the max-imum of the number of variables in the wi’s plus 2. Set M = PNK.Let w(x1, . . . , xn, z) be a commutator with n > M and w actually de-pending on x1, . . . , xn. Since M ≥ K, w(x, z) ≈ z is a law of B. ByLemma 14.12 and Lemma 14.6 w(x, z) is a sum of terms of the form

(16) r(wi(a1, . . . , ak, a0, z), z)

where r(x, z) is an idempotent binary term and where each ai is a sumof commutators and an element from the subalgebra generated by z.By Lemma 14.14 we may assume that no commutator in these sumsinvolves more than K variables. Since r is a binary idempotent termit corresponds to an element of R(U) as defined in chapter 9. Since


wi(a1, . . . , ak, a0, z) ζF z, r will distribute across sums of such elements.(This fact can be derived directly from Proposition 5.7.) Thus usingTheorem 14.15 we may assume that the ai’s are sums of an elementof the subalgebra generated by z and at most N commutators. Sincek + 2 ≤ P we see that w is a sum of terms each of which lies in asubalgebra of F generated by at most M of the variables, i.e.,

w(x, z) = u1(x, z) + · · ·+ ur(x, z)

where ui(x, z) ζ z and ui(x, z) is contained in a subalgebra generatedby at most M of the variables. Let w(x, z) involve variables in S andpossibly z. Then, since each ui is independent of at least one variablein S,

w = wδ∗S= u1δ

∗S + · · ·+ urδ

∗S

= z + · · ·+ z

= z

in F. This proves that there is a finite set of laws of A which implythat any commutator in more thanM variables is trivial. Of course thesecond part of Theorem 14.9 follows from this. The first part followsfrom this and Higman’s Lemma. �

Theorem 14.16. If A is a finite nilpotent algebra in a modularvariety V, which is a product of algebras of prime power order, then Ahas a finite basis for its laws. Moreover there is an integer M such thatif w(x, z) is a commutator in more then M variables, then A satisfiesw(x, z) ≈ z.

Proof. LetA be nilpotent of class c and suppose thatA is a directproduct of A1, . . . ,Ak, where each Ai has prime power order. Theneach Ai is nilpotent so by Theorem 14.9 there is an integerM such thatif w(x, z) is a commutator in more then M variables then w(x, z) ≈ zis a law of each Ai. Hence w(x, z) ≈ z is a law of A. The proof that Ais finitely based is by induction on the block size of [1, 1] as in the proofof Theorem 14.9. Again we choose θ ∈ Con A with 0 ≺ θ ≤ (1]c−1 ≤ ζand let C = A(θ)/∆θ,1. Let w1, . . . , wm be a set of commutators suchthat w1(x, z) ≈ z, . . . , wm(x, z) ≈ z is a basis ofA/θ×C. The only partof the proof of the previous theorem which required that A have primepower order was the proof of Theorem 14.15. However the conclusionof Theorem 14.15, with N =M , follows from the fact that

∑

S

(−1)|S|wi(x1, . . . , xj−1, yS, xj+1, . . . , xn, x0, z) ≈ z


(where the sum is over all S ⊆ {1, 2, . . . ,M+1}) is a law of A, which itis since the left side is a commutator involving more that M variables.

�

It is an open problem if every finite nilpotent algebra is finitelybased. However M. Vaughan-Lee [81] has constructed a 12 elementnilpotent loop such that there is no M such that commutators in morethen M variables are trivial. Thus Theorem 14.9 and Theorem 14.16are not true for arbitrary finite nilpotent algebras.

CHAPTER 15

Pure Lattice Congruence Identities

This chapter is being written.In Chapter 8 we examined several serveral congruence identities;

this is, equations in the language with three binary operations ∨, ∧and the commutator operation. We call such equations congruenceidentities. We examined the consequences of congruence identities inChapters 8 and 10.

In this chapter we present some of the major results about congru-ence identities that only involve join and meet and not the commutator.We concentrate on results that were proved with the aid of, or whoseproofs have been simplified by, the commutator. [Check that this istrue in the end.] We let

Con(V) = {Con(A) : A ∈ V}

and define the congruence variety associated with a variety V to bethe variety of lattices generated by all the congruence lattices of themembers of V; that is, V (Con(V)).

Theorem 15.1. Let V be a modular variety and let A be its Abeliansubvariety. Let R = R(A) be the ring associated with A and let M bethe variety of left R–modules.

(1) The congruence variety of V contains that of M.(2) If V is Abelian, that is, V = A, then congruence variety of V

equals that of M.

[better notation and terminology.]

Proof. Of course the first statement follows from the second whichfollows from Proposition 9.19 and Lemma 9.14. �

[To be covered:

• the congruence variety of an Abelian variety is that of thevariety of modules over its ring and the congruence variety ofevery modular variety contains that of its Abelian subvariety.• Describe the congruence varieties assoc with modules.• minimal modular congruence varieties• Palfy Szabo thm, no proof.

135

136 15. PURE LATTICE CONGRUENCE IDENTITIES

• Day Kiss• Freese Jonsson• Haiman• Freese Lipparini• Freese• Mal’cev condition for every lattice eq (Czedli, Lipparini, et al)• Problem: Does Con(A) have a representation as a lattice ofpermuting equivalence relations?

[Describe all congruence varieties associated with modules, Hutchin-son.]

So what are the congruence varieties of varieties of modules? Foreach function f defined on the set of prime numbers into ω∪{∞}, we de-fineRf to be the ring Z[X ]/I whereX = {xp : p a prime with f(p) 6=∞}and I is the ideal generate by the elements xpp

f(p)+1 − pf(p).For a ring R with 1 of characterisic 0, let fR(p) be the least integer

k such that there is an a ∈ R such that apk+1 = pk. If no such k exists,fR(p) = ∞. Let Zn = Z/nZ denote the ring of integers modulo n.The congruence varieties associated with varities of modules have beencharacterized by G. Hutchinson and G. Czedli [51] as follows:

Theorem 15.2. Let R be a ring with 1 and let MR be the varietyof R–modules. Let n be the characteristic of R. Then

(1) if n > 0 the congruence variety of MR is the same as the oneassociated with Zn,

(2) if n = 0 the congruence variety of MR is the same as the oneassociated with RfR.

[We may give a proof or a sketch or some exercises outlining theproof.]

One can easily check the order within the congruence varieties as-sociated with modules:

Theorem 15.3. Let n and m > 1 be integers, p a prime, andsuppose n = pek, where p does not divide k. Then

(1) V(Con(MZn)) ⊆ V(Con(MZm

)) if and only if n divides m,(2) V(Con(MZn

)) ⊆ V(Con(MRf)) if and only if e ≤ f(p),

(3) V(Con(MRf)) ⊆ V(Con(MZn

)) never holds,(4) V(Con(MRf

)) ⊆ V(Con(MRg)) if and only if f(p) ≤ g(p)

for all primes.

Using Theorems 15.1 and 15.3 we can show that the minimal mod-ular congruence varieties are just those associated with the varieties ofvector spaces over prime fields.

15. PURE LATTICE CONGRUENCE IDENTITIES 137

Corollary 15.4. If V is a congruence modular, nondistributivevariety of algebras then V(Con(V)) contains either V(Con(MZp

)), forsome prime p, or V(Con(MQ)), where Q is the rationals.1

Since, if α/β is an interval in Con(A) which contains a sublatticeisomorphic to M3 with greatest element α and least element β, it isAbelian, there was hope that every modular congruence variety arisesfrom a variety of modules. This is Problem 8.11 which was open at thetime of the first edition of this book. But this difficult problem wasrefuted by P. P. Palfy and C. Szabo in [69] and [68].

Theorem 15.5. Let V be the variety generated by Q, the 8 elementquaternion group. Then there is a lattice identity true in all lattices ofsubgroups of Abelian groups but which fails Con(FV(5)).

We will not prove this here; the proof in [68] is quite readable.This result is much more subtle than one might suspect. See [6] forfurther discussion of this problem for group varieties. Incidentally|Con(FV(5))| = 220.

Despite this there are some positive results. A. Day and E. Kiss [22]give general conditions on a variety that do guarentee its congruencevariety is the congruence variety of some variety of modules. The def-inition of R(V) is given in Definition 9.3.

Theorem 15.6. Let V be a congruence modular variety which is notcongruence distributive. If V is residually small and locally finite (orresidually small and solvable) then the congruence variety associatedwith V is the same as that of M, the variety of modules over R(V).

Proof. By R. Wille [82] and A Pixley [73] pure lattice equationholding in the congruence lattices of a variety are definable by indem-potent, weak Mal’cev conditions. This implies that the congruencevariety associated with a variety is the same as the one associated withits idempotent reduct. Thus we may assume V is idempotent.

Checking the definition of R(V) = R(V, 1) given in Defitions 9.2and 9.3, we see that it only depends on FV(u, v0)/[θ0, θ0], where θ0 =Cg(u, v0). But since V is idempotent, θ0 = 1. Thus if V is idempotentthen R(V) = R(A), where A is the Abelian subvariety of V. Of course

1This result, due to Freese [24], has an easy direct modular-lattice theoreticproof. The result has been extended in two directions. First, Day and Freese [23]showed that Polin’s variety, mentioned in Example 11.2, is contained in every non-modular congruence variety. Second, in as yet unpublished work, Paolo Lippariniand Freese show that if V is a variety which is not congruence meet semidistribu-tive, then the congruence variety associated with V contains one of the minimalmodular congruence varieties of this corollary.


the congruence variety of A is a subvariety of the congruence variety ofV. Hence by Theorem 15.1, the congruence variety of M is containedin the congruence variety of V.

Suppose β is an Abelian congruence on A ∈ V. Then by Theo-rem 9.9 the direct sum

∑

i<λM(β, zi) of the blocks is a module overR(V, λ) whose congruence lattice is isomorphic to the interval β/0.This module is of course a module over R(V) and thus β/0 is embeddedinto the congruence lattice (the submodule lattice) of anR(V)–module.

The definition of a neutral element in a modular lattice with leastand greatest element is given in Exercise 4 of Chaper 8. If u is suchan element in L then L is a subdirect product of the intervals 1/u andu/0. It is also shown there that if Con A satisfies the congruenceequation (C1) then [1, 1] is neutral. The same argument shows that[α, α] is neutral in the interval α/0.

If V is residually small it satisfies equation (C1) by Theorem 10.14.If V is solvable then, for each A ∈ V, the solvable sequence of Defi-tion 6.1 is eventually 0:

1 = [1]0 ≥ [1, 1] = [1]1 ≥ [[1, 1], [1, 1]] = [1]2 ≥ · · · ≥ [1]k = 0

for some k. By the remarks above Con A is a subdirect product ofthe intervals [1]j/[1]j+1 each of which is embedded into the congruencelattice of an R(V)–module, proving the theorem in this case.

Now suppose V is locally finite and residually small. Let A be afinitely generated (and thus finite) algebra in V and let θ ∈ Con A.Suppose [θ, θ] = θ and let α be a lower cover of θ. We claim α is neutralin the interval θ/0. Otherwise there exist β and γ ≤ θ such that

α ≤ α ∨ (β ∧ γ) < (α ∨ β) ∧ (α ∨ γ) ≤ θ

Since α ≺ θ, this implies β ∧ γ ≤ α and θ = α ∨ β = α ∨ γ and so[θ, θ] = [α ∨ β, α ∨ γ] ≤ α ∨ (β ∧ γ) = α < θ, contradicting [θ, θ] = θ.

Thus by choosing θ0 = 1 and θj+1 to be [θj , θj ] if [θj , θj ] < θj andto be any lower cover otherwise, we obtain a chain 1 = θ0 > θ1 > · · · >θk = 0 such that θj+1 is neutral in θj/0. Hence Con A is a subdirectproduct of the intervals θj/θj+1, each of which is either embeddable inthe congruence lattice of an R(V)–module or the two element lattice,which of course can also be so ebedded. Hence Con A is embeddablein the congruence lattice of an R(V)–module. �

There are several open problems some of which were presented inChapter 8.

Problem 15.7. Is the congruence variety associated with loops (orwith quasigroups) the largest modular congruence variety?

1. THE ARGUESIAN EQUATION 139

Problem 15.8. Is there a largest modular congruence variety?

Problem 15.9. Is the congruence variety associated with loops dis-tinct from the one associated with groups?

In [47] Hobby and McKenzie show that for each finite algebra Athere is a finite loop with operators such that Con A ∼= Con B. Thusthe congruence variety associated with every locally finite, congruencemodular variety is contained in the congruence variety associated withloops.

1. The Arguesian Equation

Recall from Definition 6.1 that [δ]2 = [[δ, δ], [δ, δ]].

Lemma 15.10. Let αi, βi ∈ Con A, i = 0, 1, 2 where A is in amodular variety. Let

δ = (α0 ∨ β0) ∧ (α1 ∨ β1) ∧ (α2 ∨ β2)

Then

δ =(

(α0 ◦ β0) ∩ (α1 ◦ β1) ∩ (α2 ◦ β2))

◦ [δ]2

Proof. Of course δ contains the right side of the above equation.To see the other direction we let, for congruences α and β, α ◦m β =α ◦ β ◦ α · · · with m− 1 occurrences of ◦. We shall show that if m > 2then

(1) ((α0 ◦m β0) ∩ (α1 ◦m β1) ∩ (α2 ◦m β2)) ◦ [δ]2

⊆ ((β0 ◦m−1 α0) ∩ (β1 ◦m−1 α1) ∩ (β2 ◦m−1 α2)) ◦ [δ]2

Then if λm denotes the left side of (1) then we get λm ⊆ λ2, at leastif m is even. But if m is odd λm ⊆ λm+1 ⊆ λ2 so λm ⊆ λ2 for all m ≥ 2.

The lemma follows from λm ⊆ λ2 since if 〈a, b〉 ∈ δ then 〈a, b〉 ∈(α0 ◦m β0)∩ (α1 ◦m β1)∩ (α2 ◦m β2) for some m, which is contained inthe left side of (1).

So suppose

a (α0 ◦m β0) ∩ (α1 ◦m β1) ∩ (α2 ◦m β2) b [δ]2 c

Then a = a0i αi a1i β a2i αi a3i β a4i · · · ami = b. Let d2(x, y, z) be thegeneralized Gumm difference term defined in Exercise 9 of Chapter 6.Then by that exercise we have

a = d2(a1i, a1i, a) βi d2(a2i, a1i, a) αi d2(a3i, a, a)

βi d2(a4i, a, a) αi · · · d2(ami, a, a) = d2(b, a, a) [δ]2 b


Thus

a (β0 ◦m−1 α0) ∩ (β1 ◦m−1 α1) ∩ (β2 ◦m−1 α2) d2(b, a, a) [δ]2 b [δ]2 c

proving (1) and thus the lemma. �

Theorem 15.11. If A lies in a modular variety then Con A isarguesian. That is, if αi and βi ∈ ConA, i = 0, 1, 2 then

(2) (α0∨β0)∧ (α1∨β1)∧ (α2∨β2) ≤ (α1∧ ((γ0∧ (γ1∨γ2))∨α2))∨β1

where γ0 = (α1 ∨ α2) ∧ (β1 ∨ β2) and γ1 and γ2 are defined cyclically.

Proof. By Lemma 15.10 it suffices to show that both [δ]2 and

(α0 ◦ β0) ∩ (α1 ◦ β1) ∩ (α2 ◦ β2)

lie in the right side of (2). The latter is straightforward. For the formerlet ε be the expression obtained from δ using the distributive law:

ε =(α0 ∧ α1 ∧ α2) ∨ (α0 ∧ α1 ∧ β2) ∨ (α0 ∧ β1 ∧ α2) ∨ (α0 ∧ β1 ∧ β2)

(β0 ∧ α1 ∧ α2) ∨ (β0 ∧ α1 ∧ β2) ∨ (β0 ∧ β1 ∧ α2) ∨ (β0 ∧ β1 ∧ β2)

Now[δ]2 ≤ [δ, [δ, δ]] ≤ [α0 ∨ β0, [α1 ∨ β1, α2 ∨ β2]] ≤ ε

and clearly ε is contained in the right side, proving the theorem. �

Related Literature

Reading the first treatises on general commutator theory was, tomany longtime enthusiasts of the subject of varieties, like opening ahidden door onto a bright new world of unsuspected possibilities. Itappeared that many problems long solved (or partially solved) for dis-tributive varieties and varieties of modules, could now be attacked di-rectly in modular varieties. The high expectation has been realized, inmany instances. We shall mention several areas, and papers, in whichcommutator theory has been exploited very successfully.

Smith [79] obtained new uniqueness theorems for direct decomposi-tions of finite algebras in varieties with permuting congruences. Gummand Herrmann [42] expanded these results to modular varieties and re-fined them. Two algebras A and B in a modular variety V are said tobe it isotopic in V if there is an algebra C in V and an isomorphismbetween A×C and B×C which commutes with the projections ontoC. (Exercise: If A and B are isotopic in V then C can be taken tobe an Abelian algebra in V.) Let A, B, C be algebras in a modularvariety V such that the congruences of A satisfy the ascending chaincondition, and the congruences below the center of A satisfy the de-scending chain condition. The authors proved that if A×B and A×Care isomorphic (or merely isotopic in V), then B and C are isotopic inV. They also proved an isotopic refinement theorem for algebras whichsatisfy these chain conditions. (For finite algebras, these results wereessentially proved by B. Jonsson in [53]).

The next group of applications appeared in a sequence of five papersco-authored by McKenzie. The first paper, Freese-McKenzie [29], con-tained a characterization of finitely generated, residually small, modu-lar varieties, proved here in Chapter 10, Section 3. The second, Burris-McKenzie [8] and [9],determined the structure of locally finite modularvarieties with a decidable first order theory. It turned out that thesevarieties decompose as the product of an Abelian variety and a discrim-inator variety. This pointed up the desirability of determining whichfinite rings have a decidable theory of modules, a problem that is stillopen. the third, Baldwin-McKenzie [1], solved a counting problem

141

142 RELATED LITERATURE

for modular varieties. The weak fine spectrum function of a varietyV is the function f(κ) which is defined to be ω plus the number ofnon-isomorphic κ-element members of V, κ a cardinal. The authorsproved that modular varieties of countable type have precisely six dif-ferent weak fine spectrum functions. McKenzie [64] investigated resid-ual smallness in varieties of rings. It turned out that for these varieties,(and for varieties of linear associative algebras over any commutativering K) the main theorem of the paper in this series does not requirethe hypothesis of finite generation. Thus a variety of K-algebras isresidually small if and only if it satisfies the congruence identity (C1)of chapter 8. McKenzie [63] used commutator theory to characterizethe finitely generated varieties that have only finitely many finite di-rectly indecomposable algebras, after first proving that a variety withthis property has permuting congruences.

Some of the results from Freese’s papers [26], [27], were provedhere in Chapter 10. The results of Chapter 13 are his. The proof ofthe extension of M. Vaughan-Lee’s finite basis theorem contained inChapter 14 is his work. The structural relation between the moduleassociated with an Abelian congruence and the algebra, Theorem 9.9,is his.

A sequence of important papers by E. Kiss deserve special mention.Kiss [57] proved that a modular variety has the property that principalcongruences are always complemented if and only if it decomposes asA⊗F where F is a filtral variety (and therefore distributive), and A isan Abelian variety whose ring is semisimple Artinian. This extendedthe result of C. Herrmann [46] that a variety with complemented mod-ular congruence lattices is Abelian, see exercise 11 of Chapter 5. Kiss[55] investigated the congruence extension property. He succeeded inrelating it very closely to two special properties of the commutator.(See Theorem 8.3). This allowed him to prove, for example, that alocally finite modular variety has the congruence extension property ifand only if the square of each of its subdirectly irreducible algebras hasthe property. Closely related to this work is a paper of B. A. Davey andL. G. Kovacs [18], which applies commutator theory to the study ofinjective hulls, and the existence of enough injectives, in modular vari-eties. Kiss [58] presented a short and elegant proof of a result of Burrisand McKenzie: If K is a finite class if finite algebras such that everymember of V = V (K) is a Boolean product of algebras from K,then V

is the product of an Abelian variety and a discriminator variety.In a very recent paper, [56], Emil Kiss gave a very nice general-

ization of Gumm’s difference term. Recall that we defined a differenceterm to be a term d(x, y, z) such that d(x, x, z) = z and d(x, z, z) [θ, θ] x

RELATED LITERATURE 143

whenever x θ z. Proposition 5.7 gave very useful necessary and suf-ficient conditions for [α, β] = 0 in terms of this difference term whenα and β are comparable. The hypothesis that α and β be compara-ble is necessary in that theorem. [However, see exercise 8 where it isshown that Proposition 5.7 does work in permutable varieties or if α isAbelian. See also McKenzie [65] for some related theorems using theGumm terms constructed in Theorem 6.4] Kiss defines a 4-differenceterm for V to be a term q(x, y, u, v) satisfying q(x, y, x, y) = x andq(x, x, u, u) = u, and if A ∈ V and

(∗)

[

a cb d

]

,

[

a c′

b d

]

∈ A(α, β)

(see Definition 4.7) then

q(a, b, c, d) [α, β] q(a, b, c′, d).

In a manner analogous to Theorem 5.5 and Proposition 5.7 he showsthat every modular variety possess a 4-difference term, and that it canbe used to complete diagrams as shown in the figure.

implies

a b

c d

βα γ γ ∨ (α ∧ β)

q c d

a b

βα γ

Compare this figure with the figure of Theorem 5.5. He then goeson to prove an analogue of Proposition 5.7, but without the assumptionthat the congruences are comparable. Namely, he shows that [α, β] = 0if and only if q : A(α, β)→ A is a homomorphism and if (∗) holds then

q(a, b, c, d) = q(a, b, c′, d).

In a distributive variety one can take q(x, y, u, v) = u; in a permutablevariety with Mal’cev term p(x, y, z) one can take q(x, y, u, v) = q(x, y, v).If q is a 4-difference term then d(x, y, z) = q(x, y, z, z) is a differenceterm.

In this same paper Kiss answers a question raised by McKenzie in[63]. He shows that in a finitely directly representable variety everydirectly indecomposable algebra is either finite or Abelian.

Two recent papers of C. Bergman [2] and [3] studied the congru-ence extension property in modular varieties. Building on results pre-sented there, Bergman and McKenzie have shown [unpublished] thata residually small modular variety γ with the amalgamation propertyalso possesses the congruence extension property, provided that one of

144 RELATED LITERATURE

these conditions is satisfied: (1) V is distributive; (2) V is a variety ofgroups; (3) FV(2) is finite.

The originalOverview manuscript mentioned the problem of findingring R(V, 1, 0) when V is the variety of groups or of commutative rings.This problem was solved first by David Hobby, essentially before thatmanuscript was circulated. P. Zlatos solved the problem independentlyand B. Sivak has calculated these rings for numerous varieties of groups.For the variety of groups the ring is Z[x, x−1]. Day and Kiss [22]show that for V the variety of groups, R(V, n, 0) is the group ringover the integers of the free group on n generators. L. G. Kovacs inunpublished work clarified exactly how Z[x, x−1] acts on an Abeliangroup congruence. Let H be an Abelian normal subgroup of G andlet z ∈ G. Recall that if θ is the congruence associated with H, thenM(θ, z) is the H coset containing z with addition defined by a + b =d(a, z, b) = az−1b. It is easy to see that under this addition eachcoset is isomorphic to H as an Abelian group. Kovacs showed that x(from Z[x, x−1]) acts on this coset by conjugation by z. In particular,the modules M(θ, z2) will not be isomorphic unless z−1

1 z2 lies in thecentralizer of H.

The paper of A. Day and E. Kiss [22] has already been mentionedabove and in connection with congruence identities of locally residuallysmall varieties, in Chapter 8. The paper contains some suggestiveresults that may turn out to be very significant. For one example:the ring R(V, 1, 0) of a variety is isomorphic to a ring produced viathe classical von Neumann Coordinatization process from a natural itframe of congruence in a free algebra.

W. A. Lampe, who was one of the independent discoverers of theterm condition, used it to show that certain compactly generated lat-tices can be represented as congruence lattices only by algebras witha large set of operations, see R. Freese, W. A. Lampe, and W. Taylor[28]. Recently he has solved an old problem by showing that M3 isnever the lattice of subvarieties of a variety V, Lampe [60]. V hereis not assumed to be modular. If we let F be the free algebra FV(ω)with the endomorphism adjoined as additional operations, then it iswell known that the lattice of subvarieties of V is dually isomorphicto Con F. Lampe’s proof shows that if V is modular then Con Fsatisfies [1, 1] = 1 (and hence Con F cannot have an M3 at the top).

A. Ursini has worked out a theory of it ideal determined varieties,having a constant 0 and the property that every it ideal is a congruenceclass of a unique congruence. (Examples: groups, rings, loops.) InGumm-Ursini [43], the authors proved that every ideal determined

RELATED LITERATURE 145

variety is congruence modular, and they sketched some details of atheory of commutators of ideals. This theory covers a range of varietiesin which commutator theory is more directly accessible to the intuitionthan in the general modular varieties.

To conclude, let us address the question: Is viable commutator the-ory inherently restricted to modular varieties, or is it possible to extendusable fragments of it into some wider arena? The paper P. Zlatos [83],is an attempt to provide axioms for a more general theory. In the bookby D. Hobby and R. McKenzie [47], a theory of congruences in finitealgebras is developed, called by the authors tame congruence theory,which reveals that the ‘commutator’ C(α, β), defined in Chapter 3, hasinteresting properties when restricted to locally finite algebras. If wedefine solvability in terms of this commutator, then the congruence lat-tice of a locally finite algebra A admits a congruence ∼ under whichα ∼ β (where α ≥ β) if and only if restricted to every finite subalgebraB, β|B is solvable over α|B. Moreover, (Con A)/∼ satisfies the meetsemidistributive law (x∧y = x∧ z ⇒ x∧ (y∨ z) = x∧y). If A belongsto a locally finite variety that obeys a nontrivial pure congruence iden-tity, then each equivalence class α/∼ is a modular lattice; moreover,Abelian congruences of A then have the expected structure, as if Abelonged to a modular variety. In the first paragraph of Chapter 8, weremarked that there exist nonmodular varieties which satisfy a non-trivial pure congruence identity. Among the results of McKenzie andHobby is the theorem that no locally finite variety of this kind can beresidually small.

Solutions To The Exercises

Chapter 1

1.1. A term for V has the form r1x1 + · · ·+ rnxn where ri is eitheran element of the ring or an integer. (We will usually only unitary R-modules, in which case each ri is in the ring.) From this description ofterms it is easy to see that the term condition holds for all submodulesM andN. For the second part letM andN be submodules of a moduleK. In K ×K let A = {〈m, 0〉 : m ∈ M} and B = {〈0, n〉 : n ∈ N}and let π : K × K → K be the homomorphism π(x, y) = x + y.Then π(A) = M and π(B) = N . Since A ∩ B = 0, (2) implies that[M,N] = 0.

Chapter 2

2.1. A quasigroup satisfies the cancellation law a ·b = a ·c⇒ b = c,since b = a \ (a · b) = a \ (a · c) = c. Similarly a · c = b · c ⇒ a = b.Thus the map b → a · b is one-one. Since a · (a \ b) = b, it is onto aswell. A universal algebraic proof of the last statement of the problemis this. Let A be a finite quasigroup and let F = FV (A)(x, y) be thefree algebra on two generators. Let B be the smallest subset of Fcontaining x such if b ∈ B then b · y ∈ B. B inherits the cancellationand since F is finite, B is finite. Thus the map b → b · y, b ∈ B, isone-one. Since B is finite and the image is in B, it is onto so there isa b ∈ B such that b · y = x. Of course there is a term t(x, y) usingonly · which represents b. Hence, x = t(x, y) · y. Since x = (x/y) · ywe have t(x, y) = x/y by cancellation. Notice that we have actuallyproved a little more. Namely x/y is equivalent to a term of the form(· · · (x · y) · y) · · · ) · y.

2.2. We calculate

p(x, x, y) = (x/(x \ x)) · (x \ y)

= ((x · (x \ x))/(x \ x)) · (x \ y)

= x · (x \ y)

= y

147

148 SOLUTIONS TO THE EXERCISES

Similar calculations establish the other identities.

2.3. This is established with straightforward calculations; see Day [19].

2.4. Since 〈b, d〉 ∈ γ ≤ γ ∨ (θ0 ∧ θ1), this follows from the ShiftingLemma (2.4).

2.5. Clearlymi(e, e, c, c) θ m(e, f, d, c) andmi(e, e, c, c) α2 mi(e, e, e, e) =e = mi(e, d, d, e) α2 mi(e, f, d, c). Thus since θ ∧ α2 ≤ ψ, we havemi(e, e, c, c) ψ mi(e, f, d, c). Thus we have the relations indicated inthe figure.

α1

θ ψ

mi(a, b, d, c)

mi(a, a, c, c)

mi(e, f, d, c)

mi(e, e, d, c)

Thus mi(a, b, d, c) ψ mi(a, a, c, c) by the Shifting Lemma (2.4) andso a ψ c by Lemma 2.3.

Chapter 4

4.1. Note[

p(p(x, b, y), b, b) p(p(x, b, b), b, b)p(p(b, b, y), b, x) p(p(b, b, b), b, x)

]

=

[

p(x, b, y) xp(y, b, x) x

]

is in M(θ, ψ). Thus p(x, b, y)[θ, ψ]p(y, b, x). Since [θ, ψ] = 0, x + y =y + x. Similarly

[

p(p(x, b, b), b, p(y, b, z)) p(p(x, b, y), b, p(y, y, z))p(p(x, x, b), b, p(y, b, z)) p(p(x, x, y), b, p(y, y, z))

]

=

[

p(x, b, p(y, b, z)) p(p(x, b, y), b, z)p(y, b, z) p(y, b, z)

]

shows that x+ (y + z) = (x+ y) + z.

4.2. Both are the congruence onA(ψ) generated by {〈〈x, x〉, 〈y, y〉〉 :

x θ y}. The statement

[

a bc d

]

∈ ∆ψ,θ is equivalent to 〈〈a, b〉, 〈c, d〉〉 be-

ing in this congruence which in turn is equivalent to

[

a cb d

]

∈ ∆θ,ψ.

4.3. Parts (i) and (ii) are obvious as is most of (iii). If

[

x yu v

]

∈

∆θ,ψ then

[

u vx y

]

∈ ∆θ,ψ simply because ∆θ,ψ is symmetric. Clearly if[

x yu v

]

∈ M(θ, ψ) then

[

u vx y

]

∈ M(θ, ψ). Now by Lemma 4.8, ∆θ,ψ is

CHAPTER 4 149

the transitive closure of M(θ, ψ) and it follows easily from this that theabove implication holds for ∆θ,ψ.

4.4. Since

[

a ac d

]

∈ ∆θ,ψ and

[

ab

]

∈ θ, a, b, c, and d are θ-related.

Now we have the relations indicated in the figure, where the ηi’s arethe projection kernels.

η1

η0 ∆θ,ψ

[

bc

]

[

bd

]

[

ac

]

[

ac

]

Since η0∧ η1 = 0, the result follows from the Shifting Lemma (2.4).

4.5. Let ∆ = ∆θ,ψ and suppose

[

x yu v

]

and

[

u vr s

]

are in ∆.

We will apply Lemma 2.3 with a =

[

xr

]

, b =

[

xu

]

, c =

[

ys

]

, and

d =

[

yv

]

. Now algorithmb, d〉 ∈ ∆ by assumption. Since

[

ur

]

∆

[

vs

]

and

[

uu

]

∆

[

vv

]

we have

[

mi(u, u, v, v)mi(r, u, v, s)

]

=

[[

ur

]

,

[

uu

]

,

[

vv

]

,

[

vs

]]

∆

[

ur

]

We also have[

mi(u, u, v, v)mi(r, r, s, s)

]

= mi

[[

ur

]

,

[

ur

]

,

[

vs

]

,

[

vs

]]

∆

[

ur

]

Hence by the previous exercise

[

mi(x, x, y, y)mi(r, u, v, s)

]

∆

[

mi(x, x, y, y)mi(r, r, s, s)

]

. This

says mi(a, b, d, c) ∆ mi(a, a, c, c) and so by Lemma 2.3 a ∆ c, i.e.,[

x yr s

]

∈ ∆. Thus ∆θ,ψ is a transitive relation on A(ψ) and it contains

M(θ, ψ). Hence by Lemma 4.8 ∆θ,ψ ⊆ ∆θ,ψ and by symmetry they areequal.

4.6. Let γ = (∧

αi : β). Then [γ, β] ≤∧

αi and γ is the largest suchcongruence. Since [γ, β] ≤

∧

αi ≤ αi, γ ≤∧

(αi : β). By monotonicity[∧

(αi : β), β] ≤∧

[(αi : β), β] ≤∧

αi. Hence∧

(αi : β) ≤ γ.


Chapter 5

5.1. Let α, β and γ be congruences which pairwise intersect to 0and pairwise join to 1. Then [1, 1] = [α ∨ β, α ∨ γ] = [α, α] ∨ [α, γ] ∨[β, α] ∨ [β, γ] ≤ α ∨ (β ∧ γ) = α. Similarly [1, 1] ≤ β and so [1, 1] = 0.

5.2. Clearly (ii) ⇒ (i) ⇒ (iii). Let B is a subdirect power of twocopies of A such that Con B has M3 as a 0, 1-sublattice. By theprevious exercise B is Abelian. A is a homomorphic image of B, soby Proposition 4.4(1) Ais Abelian. Hence (iii) ⇒(iv). Suppose that Ais Abelian and let ∆1,1 be the congruence in Definition 4.7. If followsdirectly from Theorem 4.9 (and Exercise 4.3(iii)) that ηi ∧ ∆1,1 = 0,where ηi, i = 0, 1 are the projection kernels. Moreover it is shown inthe proof of Theorem 4.11 that ηi ∨∆11 = 1. Hence (iv) ⇒ (ii).

5.3. By Exercise 2, B is Abelian, and so affine by Herrmann’stheorem. Thus x − y + z is a term operation on B. In particular, Bhas a Mal’cev term and thus has permutable congruences. Hence theprojection kernels permute and join to 1 so the subdirect product isdirect.

5.4. Suppose that

[

xy

]

γ ◦ α

[

uv

]

. Then there is a b ∈ B such that[

xy

]

γ

[

ub

]

α

[

uv

]

. Then we have

α

βγ

[

uv

]

[

xv

]

[

ub

]

[

xb

][

xy

][

xd(y, b, v)

]

Hence

[

xy

]

α

[

xd(y, b, v)

]

γ

[

uv

]

, showing that α and γ permute.

5.5. The inverse map is y 7→ d(y, v, u). To see this note that byProposition 5.7

d(d(x, u, v), v, u) = d(d(x, u, v), d(u, u, v), d(u, u, u))

= d(d(x, u, u), d(u, u, u), d(v, v, u))

= d(x, u, u) = x.

It follows that the map is a bijection. Another application of Proposi-tion 5.7 shows that it preserves addition.

CHAPTER 5 151

5.6. To see that (i) holds apply condition (iii) with x = y, z = u =u′, α = γ = 0, and β = 1. This gives d(u, u, y) = y. Now by takingu = u′ it is easy to derive the Shifting Lemma from condition (iii) (cf.Exercise 4 of Chapter 2). As pointed out in Chapter 2, this implies Vis modular. Now suppose that a θ b. Apply (iii) to A(θ) with x = z =[

bb

]

, y = u′ =

[

ab

]

, u =

[

aa

]

, α = η1, β = η0, and γ = ∆θ,θ (the ηi’s are

the projection kernels). This gives

[

ad(a, b, b)

]

=

[

d(a, a, a)d(a, b, b)

]

∆

[

bb

]

,

which by Theorem 4.9 implies a [θ, θ] d(a, b, b), as desired.

5.7.

[

(x · y)/(y · x) (1 · y)/(y · 1)(x · 1)/(1 · x) (1 · 1)/(1 · 1)

]

=

[

(x · y)/(y · x) 11 1

]

∈ M(1, 1).

Since A is Abelian, this gives (x · y)/(y · x) = 1 and so x · y = y · x.Starting with ((x ·y) ·z)/(x · (y ·z)) in the upper left corner of a matrix,a similar proof gives associativity.

5.8. Let p(x, y, z) be either of the Mal’cev terms given in Exercise 2of Chapter 2. Then p is a difference term and by Proposition 5.7it suffices to show that p commutes with the basic operations. ByExercise 1 of Chapter 2 / and \ are equal on G to a term using onlythe multiplication. Thus it is enough to show that p commutes withmultiplication, i.e.,

(∗) p(x1, y1, z1) · p(x2, y2, z2) = p(x1 · x2, y1 · y2, z1 · z2).

Since G has 4 elements, there are only 46 = 4096 cases to check. Youmay want to use a computer.

There is an easier solution to this problem. If A is an affine algebrawith A = 〈A,+,−, 0〉 as its Abelian group, then it is not hard toshow that if t(x, y) is a binary term operation on A then there are

endomorphisms σ and τ of A and c ∈ A such that t(x, y) = σ(x) +τ(y)+c, see Chapter 9. Knowing what to look for, it is not hard to findsuch a representation for the multiplication in G. Namely start withthe four element group which is the direct product of two copies of thetwo element group, i.e., {0, 1, 2.3} with x + x = 0, and x + y = z if xand y are distinct members of {1, 2, 3} and z is the remaining member.Let σ be the identity, τ transpose 2 and 3, and c = 3. Since τ is anautomorphism of the group, the operation x · y = x+ τ(y) + 3 definesan Abelian quasigroup which equals G.

5.9. Take the cyclic group of order four, {0, 1, 2, 3}, and let σinterchange 2 and 3. Then the multiplication of the given quasigroup isx·y = σ(x)+y. Since σ is not a group automorphism, A is not Abelian.In fact (∗) of the previous solution fails with y1 = 1 and all other


variables 0. However, it is possible to show that p(x, y, z) = x− y + zand thus x+c y = p(x, c, y) is an Abelian group operation.

5.10. It is routine to verify that G is a loop with 〈0, 0〉 as itsidentity element and that the second projection is a homomorphism.It is also easy to see that H ∼= Z4. Suppose that [θ, θ] = 0. Then forall z ∈ H, and all x, y, y′ with yθy′

(∗) (x+ (y + z))/(x+ y) = (x+ (y′ + z))/(x+ y′).

To see this just note that the following matrix is in M(θ, θ).[

(x+ (y + z))/(x+ y) (x+ (y + 0))/(x+ y)(x+ (y′ + z))/(x+ y) (x+ (y′ + 0))/(x+ y′)

]

=

[

(x+ (y + z))/(x+ y) 0(x+ (y′ + z))/(x+ y′) 0

]

Now if we let z = 〈1, 0〉, x = 〈0, 1〉, y = 〈1, 1〉, and y′ = 〈2, 1〉 thenthe left side of (∗) evaluates to 〈2, 2〉/〈0, 2〉 = 〈2, 0〉 and the rightside evaluates to 〈3, 2〉/〈2, 2〉 = 〈1, 0〉. Thus (∗) fails in G and hence[θ, θ] 6= 0.

5.11. Since the congruence lattices are complemented, every sub-directly irreducible algebra in V is simple. If all of these subdirectlyirreducible algebras are Abelian then V is Abelian. Thus let A be asimple, non-Abelian algebra in V. Let I be an infinite set and let Bthe direct product of |I| copies of A. Define λ ∈ Con B by xλy ifand only if xi = yi for all but finitely many i. Let λ′ be a complementof λ in Con B, and let ηi be the kernel of the projection onto the ith

coordinate. Clearly ηi 6≥ λ. If λ′ ≤ ηi for all i, then λ′ = 0. But then

λ = 1, which is false. Thus, for some i, λ′ 6≤ etai. From this and thefact that ηi is a coatom it follows that ηi ∧ (λ ∨ λ′), ηi, ηi ∧ (λ′ ∨ λ)generates an M3 with greatest element 1. But then [1, 1] ≤ ηi, whichimplies A is Abelian by Proposition 4.4(1). This contradiction provesthat V is Abelian.

Chapter 6

6.1. By reversing the roles of α and β in Theorem 6.2 we obtainβ◦α ⊆ [β]k◦α◦β. Now if 〈x, y〉 ∈ α◦β then 〈y, x〉 ∈ β◦α ⊆ [β]k◦α◦β.Hence 〈x, y〉 ∈ β ◦ α ◦ [β]k, as desired.

6.2. By the last exercise α ◦ β ⊆ β ◦ α ◦ [β]n. By Theorem 6.2α ◦ [β]n ⊆ [α]m ◦ [β]n ◦ α and hence α ◦ β ⊆ β ◦ [α]m ◦ [β]n ◦ α. Ofcourse if [α]m and [β]n permute for some m and n, then α ◦ β ⊆ β ◦ αas [α]m ⊆ α and [β]n ⊆ β.

CHAPTER 6 153

6.3. Clearly it suffices to show that if α and β permute then [α, α]and β permute. Suppose that x [α, α] y β z. Then since α and βpermute there is some u such that x β u α z. Thus we have therelations indicated in the figure.

β

α [α, α]

u

z

x

y

By the shifting lemma u [α, α] ∨ (α ∧ β) z. We will show that[α, α] and α∧ β permute. From this it follows that for some v we havex β u α ∧ β v [α, α] z, i.e., x β v [α, α] z, showing that [α, α] and βpermute. To see that [α, α] and α∧β permute observe by the previousexercise, with n = m = 1,

(α ∧ β) ◦ [α, α] ⊆ [α, α] ◦ [α, β]1 ◦ [[α, α]]1 ◦ (α ∧ β)

= [α, α] ◦ (α ∧ β)

since both [α ∧ β]1 and [[α, α]]1 are contained in [α, α].

6.4. Suppose that (α, α]m = 0 = (β, β]n. We claim that (α∨β, α∨β]m+n = 0. Now (α ∨ β, α ∨ β]m+n is the commutator of m + n + 1copies of α ∨ β, associated right to left. If we expand this using thecommutator, we obtain a sum of terms of the form

[γ0, [γ1, . . . [γm+n−1, γm+n] . . .],

where each γi is either α or β. Now either there are at least m+ 1 γ’sequal to α or n+1 γ’s equal to β. Thus each term in our sum is eitherless then or equal to (α]m or (β]n, and hence 0.

6.5. Let x, y ∈ A ∈ V and suppose that x θ y. We must show thatx [θ, θ] p(x, y, y). Since p(x, y, y) = qn(x, y, y), and q0 = x, it suffices toshow that, for each i, qi(x, y, y) [θ, θ] qi+1(x, y, y). For i even the twosides are actually equal by Theorem 6.4(3). For i odd first we need toshow that for all i, qi(x, y, y) [θ, θ] qi(x, x, y). The following matrix isin M(θ, θ):

[

qi(x, y, y) qi(x, x, y)qi(x, y, x) qi(x, x, x)

]

=

[

qi(x, y, y) qi(x, x, y)x x

]

.

Hence qi(x, y, y) [θ, θ] qi(x, x, y), as desired. Now for i odd, qi(x, y, y) [θ, θ]qi(x, x, y) = qi+1(x, x, y) [θ, θ] qi+1(x, y, y), proving the result.

6.6. Each of (1) to (8) is either obvious or can be verified bystraightforward calculations. For example, to see that (3) holds, sup-pose that K(a, b, c, d) holds and that s(c, a, e) = s(c, b, e). Let u =


s(d, a, e), v = s(d, b, e), and w = s(c, a, e) = s(c, b, e). We need toshow that u = v. Using K(a, b, c, d) (with the s(a, e) = s(a, e0, . . . , ek)from the definition of K(a, b, c, d) taken to be s(e0, a, e1, . . . , ek) fromthe present context) we have

qi(u, w, v) = qi(s(d, a, e), s(c, b, e), s(d, b, e))

= qi(s(d, a, e), s(d, b, e), s(d, b, e))

= qi(u, v, v).

Similarly qi(u, w, v) = qi(u, u, v). Hence qi(u, v, v) = qi(u, u, v). Thus,by Theorem 6.4, u = p(u, v, v). But K(a, b, c, d) also implies thatp(u, v, v) = p(w,w, v) = v. Hence u = v, as desired.

To see (4) use (2) and (3) repeatedly: H(a, b, c, d) → K(a, b, c, d)→ H(c, d, a, b)→ K(c, d, a, b)→ H(a, b, c, d).

Let θ = Cg(a, b) and ψ = Cg(c, d) and assume that K(a, b, c, d)holds. Then it follows from (4) to (8) thatK(a, b) = {〈x, y〉 : K(a, b, x, y)}is a congruence and 〈c, d〉 is in this congruence. Hence ψ ⊆ K(a, b) andso if 〈u, v〉 ∈ ψ then K(a, b, u, v) holds. Repeating this argument weget thatK(x, y, u, v) [and hence H(x, y, u, v)] holds whenever 〈x, y〉 ∈ θand 〈u, v〉 ∈ ψ. We want to show that C(a, b, c, d) holds, i.e., [θ, ψ] = 0.We show this by showing the θ, ψ term condition holds. So let t be aterm operation and xi θ yi, i = 0, . . . , m−1 and ui ψ vj , j = 0, . . . , k−1.

Then

[

t(x,u) t(x,v)t(y,u) t(y,v)

]

∈ M(θ, ψ). Suppose t(x,u) = t(x,v). We want

t(y,u) = t(y,v). First observe that we may assume that m = 1, i.e.,x = (x) and y = (y), since the general case of the term conditioncan be derived from this. Let w = t(x,u) = t(x,v) and u = t(y,u)and v = t(y,v). Just as in the argument above, it will suffice toshow that qi(u, u, v) = qi(u, w.v) = qi(u, v, v), for i = 0, . . . , n, andp(u, v, v) = p(w,w, v). By Theorem 6.4(2), we have

qi(t(y,u), t(x,u), t(y,u)) = qi(t(y,u), t(y,u), t(y,u))

By the symmetry in (4) we have H(uj, vj, x, y), j = 0, . . . , k − 1. Ap-plying this k times to (the last u in) the above equation we obtain:

qi(t(y,u), t(x,u), t(y,v)) = qi(t(y,u), t(y,u), t(y,v)).

Thus qi(u, w, v) = qi(u, u, v). Similarly qi(u, w, v) = qi(u, v, v) andp(w,w, v) = p(u, v, v), as desired. Hence [θ, ψ] = 0, i.e., K(a, b, c, d)→C(a, b, c, d). Now by (1) they are equivalent.

6.7. By Gumm’s result, in every case we have that p(x, y, y) =x whenever 〈x, y〉 ∈ α. Let D = 〈x, y, z〉 : x α y β z ⊆ A3. First,suppose that p : D → A is a homomorphism. We wish to prove that

CHAPTER 6 155

[α, β] = 0. By the result of the previous exercise, it will suffice toprove that H(a, b, c, d) holds for all 〈a, b〉 ∈ α and 〈c, d〉 ∈ β. To dothis, suppose that s(x, y, z) is a k + 2-ary term, that e ∈ Ak, andthat s(a, c, e) = s(a, d, e). We apply our assumption to the elements〈b, a, a〉, 〈c, c, d〉, 〈ei, ei, ei〉 of D. Thus we get

p(s(b, c, e), s(a, c, e), s(a, d, e))

= s(p(b, a, a), p(c, c, d), . . . , p(ei, ei, ei), . . .)

= s(b, d, e).

Since s(b, c, e) = s(b, d, e), the displayed equations yield s(b, c, e) =s(b, d, e) as desired. This completes the proof that [α, β] = 0 if p : D→A is a homomorphism.

Now suppose that [α, β] = 0. Thus C(α, β; 0) (see Definition 3.2).To prove that p|D is a homomorphism let f be a basic k-ary operationof A and let x α y β z with x, y, z ∈ Ak. It must be show thatp(f(x), f(y), f(z)) = f(p(x,y, z)) where p(x,y, z) denotes the obviousk-tuple of elements of A. This desired equation can be rewritten as

p(f(x), f(y), f(p(y,y, z))) = p(f(y), f(y), f(p(x,y, z))).

If we replace the zi by yi, we do obtain a true equation, with both sidesequal to f(x). The desired equation now follows from C(β, α; 0).

6.8. Let C = A × B and η0, η1 be the projection congruences onC, where each of A and B has permuting congruences. According toExercise 4 in Chapter 5, ηi permutes with all congruences of C. Notethat whenever a congruence β permutes with each of δ0 and δ1 then βpermutes with δ0 + δ1. Now suppose that β permutes with [α, η0] andwith [α, η1]. Then by Theorem 6.2,

α ◦ β ⊆ [α, α] ◦ β ◦ α

⊆ ([α, η0] ∨ [α, η1]) ◦ β ◦ α

= β ◦ ([α, η0] ∨ [α, η1]) ◦ α

= β ◦ α,

so that β permutes with α. By two applications of this observationwe reduce our work to showing that congruences permute if each ofthem lies below one of the ηi. Of course, two congruences lying aboveηi (for fixed i) must permute, since the congruences of C/ηi permute.Consider first the case α ∨ β ≤ η0. The congruences α ∨ η1 = α ◦ η1


and β ∨ η1 = β ◦ η1 permute. Thus

α ◦ β ⊆ (β ∨ η1) ◦ (α ∨ η1)

= β ◦ η1 ◦ α ◦ η1

= β ◦ α ◦ η1.

giving

α ◦ β ⊆ η0 ∩ ((β ◦ α) ∩ η0) ◦ η1)

= β ◦ α.

Consider now the case α ≤ η0, β ≤ η1. Note that α = α ∨ (η0 ∩ β) =η0 ∩ (β ∨ α). Note also that α ◦ β ⊆ η0 ◦ β = β ◦ η0. Thus we have

α ◦ β ⊆ β ◦ (η0 ∩ (β ∨ α)) = β ◦ α.

this completes the proof that all congruences of C permute.

Chapter 7

7.1. Let B = {0, 1, 2} with addition modulo 3. Define a transferfunction T : B2 → B for addition by T (1, 1) = 1 and T (x, y) = 0for all 〈x, y〉 6= 〈1, 1〉. Let A = B ⊗T B. It is easy to see that Ais a loop with identity 〈0, 0〉. Let η be the kernel of the projectiononto the second coordinate. As in the proof of Corollary 7.2, η is acongruence such that η ≤ ζA and A/η is Abelian. Since B is simple,either ζA = η or ζA = 1. In the latter case A would be Abelian. To seethat this cannot happen, Let x · y denote the operation on A, so that〈x0, x1〉·〈y0, y1〉 = 〈x0+y0, x1+y1〉 unless x1 = y1 = 1 in which case it is〈x0+y0+1, x1+y1〉. Let p(x, y, z) = x·(y\z). Then p is a Mal’cev termfor A and if we let x = x′ = 〈0, 1〉, y = y′ = 〈0, 0〉, and z = z′ = 〈0, 2〉then p(x·x′, y ·y′, z ·z′) = 〈1, 0〉 but p(x, y, z)·p(x′, y′, z′) = 〈0, 0〉. HenceA is not Abelian and thus η = ζA, which implies that A/ζA ∼= Z/3, asdesired.

7.2. Let A be a nilpotent loop having four elements. If it is notAbelian then by Corollary 7.5 it must be class two and by Corollary 7.2it must have the formB⊗TC whereB andC are two element loops. Upto isomorphism there is only one loop of order 2, the integers modulo2. Thus B ∼= C. Now B⊗T B must have an identity element and thismust be either 〈0, 0〉 or 〈1, 0〉. For the former to be the identity wemust have T (x, y) = 0 if either x or y is 0. However one can easilycheck that if T (1, 1) = 1 then B ⊗T B is isomorphic to Z/4 and ofcourse if T (1, 1) = 0 then it is isomorphic to Z/2 × Z/2. A similarargument works when 〈1, 0〉 is the identity.

CHAPTER 7 157

Showing that an arbitrary loop of order four is Abelian involveslooking at the various cases for the multiplication table but is notdifficult.

7.3. Let θ0 = θ and θk+1 = [1, θk]. Since A is nilpotent and θ 6= 0,there is a positive integer r such that θr = 0 and θr−1 6= 0. Then[1, θr−1] = θr = 0. Hence 0 < θr−1 ≤ θ ∧ ζ .

7.4. We prove this by induction on |A|. Let θ = ζ ∩B2 ∈ Con B.Since B is nilpotent, θ is uniform by Corollary 7.5. Since B/θ is asubalgebra of A/ζ , |B/θ| divides |A/ζ | by induction. Now let b ∈ B.By the uniformity |B/θ| = |B|/|b/θ| and |A/ζ | = |A|/|b/ζ |. Hence|b/ζ||b/θ|B| divides |A|. But b/ζ is a ternary group and b/θ is a ternary

subgroup. Thus |b/θ| divides |b/ζ |. Hence |B| divides |A|.

7.5. Let Z = 1/ζ and choose b ∈ A− Z. Let B be the subalgebragenerated by b. Then |B| is either 1 or p or p2. The first case isout because 1 is the only idempotent element of a loop. Suppose that|B| = p. Since B is a subalgebra of A, B is nilpotent. Since |B| = p,B is an Abelian loop. By Exercise 7 of Chapter 5, B is a cyclic groupof order p. Since B ∩ Z = 1 and every element of Z commutes andassociates with all elements ofA, it is not hard to see that every elementof A has the form bic for some i, 0 ≤ i < p, and c ∈ Z. Using thisit follows that every element of A commutes and associates with allelements of A. This implies A is Abelian, contradicting |A/ζ | = p.Thus we must have |B| = p2, i.e., A = B and so A is generated by b.

7.6. Using the first relation of Lemma 7.3 we have

q(a, b, b) = fn(a, b, b) = fn(p(a, b, b), b, b) = a

and

q(b, b, a) = fn(b, a, b) = fn(p(a, a, b), a, b) = a.

7.7. Clearly p is a Mal’cev term for A. Let θ be the equivalencerelation on A with two blocks, G and H . It is easy to verify that θis a congruence relation on A and that A/θ is isomorphic to the two-element ternary group. Thus [1, 1] ≤ θ. By Proposition 5.7 to showthat [θ, θ] = 0 we need to show that p commutes with itself on theblock of θ. Since p(x, y, z) = x − y + z on these blocks, this is clear.Hence A is solvable.

Now it is easy to use this algebra to construct a solvable algebrawhich fails to satisfy the conclusions of Lemma 7.3-Corollary 7.7. ForLemma 7.6, if a, b ∈ G and c ∈ H with a 6= b then Cg(a, b) 6= 0 butCg(p(a, b, c), c) = Cg(c, c) = 0. For Corollary 7.7, let β ∈ Con A be


defined by 〈x, y〉 ∈ β if x, y ∈ G or x = y. Then β and θ have a blockin common but are not equal. So A does not have regular congruences.

7.8. Assume that the relation holds for n and let x′ = fn(p(x, b, c), b, c)so that x′ (1]n x by assumption. Then

fn+1(p(x, b, c), b, c) = p(b, p(b, p(x, b, c), p(x′, b, c)), x′).

Now the following matrix is in M((1]n, 1)[

p(b, p(b, p(x, b, c), p(x′, b, c)), x′) p(b, p(b, p(x, x, b), p(x′, b, b)), x′)p(b, p(b, p(x, b, c), p(x, b, c), x) p(b, p(b, p(x, x, b), p(x, b, b)), x)

]

=

[

p(b, p(b, p(x, b, c), p(x′, b, c)), x′) bb b

]

Thus by Definition 3.2(2), fn+1(p(x, b, c), b, c) (1]n+1 x, as desired.

7.9. If α and β are congruences on A which have a common blockthen α ∨ β and β have the same common block. So if A is not reg-ular then it has congruences α > β which share a block. In A/β thecongruence α/β is nontrivial but has a block with one element. Thiscontradicts the hypothesis that all the homomorphic images of A haveuniform congruences.

Chapter 8

8.1. Let [α, β] = α ∧ β for all congruence α and β of A. Then forcongruences χ, φ, ψ we have χ∧ (φ∨ψ) = [χ, φ∨ψ] = [χ, φ]∨ [χ, ψ] =(χ ∧ φ) ∨ (χ ∧ ψ) showing that Con A is a distributive lattice.

8.2. It suffices to prove the result for subdirect product of twofactors. So suppose that A and B are neutral and that C ⊆ A × Bwith p0(C) = A and p1(C) = B. Let ηi (i = 0, 1) be the kernels of pi.Let ν, µ be any two congruences of C. Using the assumption about Aand B and Remark 4.6, we find that [ν, µ] ∨ ηi = (ν ∨ ηi) ∧ (µ ∨ ηi),implying that ν ∧ µ ≤ [ν, µ] ∨ ηi and so

ν ∧ µ = [ν, µ] ∨ (ν ∧ µ ∧ ηi).

This is true also when we replace ν by ν ∧ ηi and mu by µ ∧ ηi. Thuswe obtain

ν ∧ µ ∧ ηi = [ν ∧ ηi, µ ∧ ηi] ∨ (ν ∧ ηi ∧ µ ∧ ηi ∧ η1−i)

= [ν ∧ ηi, µ ∧ ηi].

Putting the two displayed equations together gives ν ∧ µ = [ν, µ], asdesired.

If V = V0 ∨V1, where V is modular and V0 and V1 are distributive,then FV(3) is a subdirect product of FV0

(3) and FV1(3). Hence FV(3)

CHAPTER 8 159

is neutral and so by the previous exercise, Con FV(3) is distributive.Now by Jonsson’s Theorem 2.1, V is distributive.

8.3. The exercise is to prove that a variety V is distributive if andonly if it satisfies (C3) if and only if its finite subdirect products possessno skew congruences. According to the result of Exercise 1, V |= (C3)implies V is distributive. If V is distributive and C is a subdirectproduct of A and B in V with projection congruences ηi, and if ψ isany congruence of C, then ψ = ψ∨ (η0∧η1) = (ψ∨η0)∧ (ψ∨η1); henceψ is not skew. Finally, suppose that V has no skew congruence on finitesubdirect products. We shall show that this implies V |= (C3). If (C3)fails, then there is A in V with a nonzero Abelian congruence β. Thealgebra A(β) (see Definition 4.7 and the proof of Theorem 4.10) hascongruences ∆β,β, η0, η1 forming an M3. Thus A(β) is the subdirectproduct of A(β)/η0 and A(β)/η1, and it is clear that ∆β,β is skew.

8.4. Suppose that a is an element in a modular lattice L satisfyinga ∧ (x ∨ y) = (a ∧ x) ∨ (a ∧ y) for all elements x and y. By joining xto both sides of the equation we obtain (a ∨ x) ∧ (x∨ y) = x∨ (a∧ y).Then meet with y we obtain (a ∨ x) ∧ y = (x ∧ y) ∨ (a ∧ y). Finally,joining with a we obtain (a ∨ x) ∧ (a ∨ y) = a ∨ (x ∧ y). The proof ofthe equivalence of the conditions a∧ (x∨ y) = (a∧ x) ∨ (a∧ y) (for allx, y) and (a∨ x) ∧ (a∨ y) = a∨ (x ∧ y) (for all x, y) is now concludedby dualizing the above argument. Now if a satisfies these equationsthen it is clear that x 7→ 〈a∨ x, a ∧ x〉 is a homomorphism of L onto asubdirect product of the interval lattice 1/a and a/0. To see that thefunction is injective, suppose that a ∨ x = a ∨ y and a ∧ x = a ∧ y.Then x = x∧ (a∨ x)∧ (a∨ y) = x∧ (a∨ (x∧ y)) = (x∧ a)∨ (x∧ y) =(y ∧ a) ∨ (x ∧ y) ≤ y, and the same argument shows that y ≤ x, sox = y. Now suppose that Con A |= (C1). According to the discussionpreceding Theorem 8.1, the element a = [1A, 1A] of Con A satisfiesthe above conditions, i.e. is neutral. Thus Con A is embeddablesubdirectly into the product of the interval lattices 1/a and a/0.

8.5. Let Con A |= (C1). Suppose that α and β are Abeliancongruences of A. Observe that [α, β] ≤ [α∨β, α∨β], so that by (C1)[α, β] = [[α, β], α ∨ β] = [[α, β], α] ∨ [[α, β], β] ≤ [α, α] ∨ [β, β] = 0A.Thus α, β Abelian implies [α, α] = [α, β] = [β, β] = 0A. Now letα =

∨

{αi : i ∈ I} where {αi : i ∈ I} is the set of all the Abeliancongruences of A. Since the commutator is completely additive and,as we just showed, [αi, αj] = 0A for all i and j then it follows readilythat α is Abelian. Clearly α is the largest Abelian congruence of A.

8.6. We prove the harder of the two statements of the exercise. Letε be a congruence equation such that V |= (Ci) implies V |= ε (i = 3, 4)


for all modular varieties V. We are to show that this implication is validalso for i = 2. So let V be a modular variety that satisfies (C2). Let Abe the variety of Abelian algebras in V. Choose any algebraA in V, andset L = Con A. Let u = [1A, 1A]. Now we also have that V |= (C1) (infact, (C2) implies (C1)). So by the result of Exercise 4, L is a subdirectof the intervals 1/u and u/0. It is easy to see that this subdirectrepresentation preserves commutators. Hence the commutation lattice〈L, x∧y, x∨y, [x, y], 0, 1〉 is a subdirect product of commutation latticesL0 and L1 whose universes are the intervals 1/u and u/0, respectively.To see that L |= ε, which is the point of the exercise, one is thusreduced to a consideration of the commutation lattices Li. Now L0 isnaturally isomorphic to the commutation lattice of the algebra A/u,which belongs to A. Since A |= (C4), it follows that L0 |= ε. On theother hand, L1 = 〈u/0, x ∧ y, x ∨ y, x ∧ y〉, a distributive lattice withtrivial commutator (equal to the meet). The proof can be concludedby showing that L1 can be embedded into the commutation lattice ofan algebra in a distributive variety (in fact we can choose the variety oflattices), for this will imply that L1 |= ε. This is very easy to do. Wecan ignore the commutator since it is trivial, and regard L1 simply asa distributive lattice with 0 and 1. L1 is a sublattice of some Booleanlattice, which in turn is isomorphic to a sublattice of its own congruencelattice.

8.7. Let R be a ring with unit. For any ideal J of R, J · R =R · J = J, due to the presence of a unit element. Equivalently, for anycongruence θ of R, [θ, 1R] = θ.

8.8. Suppose that L = Con A satisfies (C8) and α, β ∈ L satisfyα ∨ β = 1, α ∧ β = 0, α ◦ β = β ◦ α. Then the natural homomorphismA to A/α×A/β is an isomorphism. Thus we can assume that α andβ are the projection congruences on a product algebra A = C × D.Now by the proof in Theorem 8.5 that (1) ⇒ (5), it follows that αand β are neutral in L and that π(λ) = 〈α ∨ λ, β ∨ λ〉 defines anisomorphism between L and (1/α)× (1/β), where the interval lattices1/α and 1/β are complete and isomorphic to Con C and Con D,respectively. From this it follows that f(

∧

{λt : t ∈ T}) =∧

{f(λt) :t ∈ T} for every family {λt : t ∈ T} ⊆ L, and this implies that fis a complete endomorphism of L, as asserted. (Note that f triviallypreserves infinite joins.)

8.9. The algebra A is actually simple and non-Abelian, from whichits neutrality follows trivially. To see this, let θ be any nonzero con-gruence of A. We can choose a0 6= 0 with 〈0, a0〉 ∈ θ, then choose k−1more elements forming a vector space basis a0, a1, . . . , ak−1 of V. Now

CHAPTER 9 161

for all x, y ∈ A.

x · y = f(a0, . . . , ak−1, x, y)

≡ f(0, a1, . . . , ak−1, x, y) (mod θ)

= 0

= g(0, a1, . . . , ak−1, x, y)

≡ g(a0, a1, . . . , ak−1, x, y) (mod θ)

= x⊕ y.

The calculation shows that x·y ≡ x⊕y (mod θ), implying that 〈x, y〉 ∈θ. Thus θ is 1A. The calculation also displays the fact that the latticeoperations x · y and x ⊕ y are polynomial operations of A, and thisimplies A is non-Abelian. Now to show that every algebra in V (A)generated by k − 1 or fewer elements is Abelian, it suffices to provethis for the subalgebras of A that are generated by k − 1 or fewerelements, because the free algebra F = FV (A)(k − 1) is a subdirectproduct of these algebras. So let B be the subalgebra of A generatedby elements b0, . . . , bk−2. Let W be the vector subspace of V subspaceby b0, . . . , bk−2. On W the operations f and g are constant, givingonly the value 0. (This follows from the definition of f and g.) Thus,clearly, W is a subuniverse of A, and we must have W = B. Then〈B,+,−, f, g〉 = 〈B,+,−, 0, 0〉 and so B is Abelian.

8.10. Every subgroup of an Abelian group is a normal subgroup.That Abelian groups have the congruence extension property followseasily from this. Now suppose that V is a variety of groups in whichcongruences can always be extended. Thus, by Theorem 8.4, wheneverA is a group in V and B, H, K are subgroups of A with H and Knormal in A, then [H∩B,K∩B] = [H,K]∩B. Suppose that A is non-Abelian. Then there exists a nontrivial cyclic subgroup B ⊆ [A,A].Taking H = K = A in the above formula leads to an immediatecontradiction.

Chapter 9

9.1.. Suppose A is Abelian and R = R(V (A)) is commutative.Using equation (10) we see that x·y = rx+sy+c for some r, s ∈ R andc ∈ A. Then (x·y)·(u ·v) = (rx+sy+c)·(ru+sv+c) = r2x+rsy+rc+sry+s2v+sc+c. Also (x ·y) ·∗y ·v) = r2x+rsy+rc+sry+s2v+sc+c,and so (x · y) · (u · v) = (x · u) · y · v) follows from the commutativityof R.

Now suppose that the identity holds. It says that · commutes withitself. From this we can show that each of ·, /, and \ commute with


each other and with themselves. For example, to see that / commuteswith itself we calculate

x = [((x/y)/(u/v)) · (u/v)] · [(y/v) · v]

= [((x/y)/(u/v)) · (y/v)] · [(u/v) · v]

= [((x/y)/(u/v)) · (y/v)] · u.

Hence ((x/y)/(u/v)) · (y/v) = x/u. But [(x/u)/(y/v)] · (y/v) = x/ualso. Thus (x/y)/(u/v) = (x/u)/(y/v), as desired. To see that · and /commute, calculate

[(x/y) · (u/v)] · (y · v) = [(x/y) · y] · [(u/v) · v] = x · u

Since [(x·u)]/(y·v)]·(y·v) = x·u also, we have (x\y)·(u\) = (x·u)/(y·v).Which shows · and \ commute.

Thus the basic operations and hence all term operations commutewith each other. By Proposition 5.7,A is Abelian. If r, s ∈ R then rs =r(s(u, v), v) = r(s(u, v), s(v, v)) = s(r(u, v), r(v, v)) = s(r(u, v), v) =sr.

To see the second statement first suppose that V (A) is permutableand every basic operation of A commutes with every idempotent termoperation of A. Then every idempotent term operation commuteswith every other idempotent term operation. By Proposition 5.7, A isAbelian and since the elements of R(V (A)) are idempotent terms, itis commutative as above.

Now suppose that A is Abelian andR = R(V (A)) is commutative.Choose an arbitrary element 0 in A and letR act onA by r·a = r(a, 0),as usual. If f is an n-ary term operation on A then since A is Abelianthere are elements rf1, . . . , rfn ∈ R and cf ∈ A

(∗) f(x1, . . . , xn) = rf1 · x1 + · · ·+ rfn · xn + cf .

If g is an m-ary term operation then it has a similar representation

g(x1, . . . , xm) = rg1 · x1 + · · ·+ rgm · xm + cg.

If one writes out both f(g(x11, . . . , x1m), . . . , g(xn1, . . . , xnm)) and alsog(f(x11, . . . , xn1), . . . , f(x1m, . . . , xnm)) in terms of the above represen-tations and uses the fact that R is commutative, one sees that f andg commute if and only if

(∗∗) (rf1 + · · ·+ rfn − 1) · cg = (rg1 + · · ·+ rgm − 1) · cf

holds in A. Now suppose that f is idempotent. Then it follows fromthe representation (∗) with x1 = · · · = xn = 0, that cf = 0. If we letx1 = · · · = xn = x in (∗) we see that (rf1 + · · · + rfn) · x = x for all

CHAPTER 9 163

x ∈ A. Now by (∗∗) it is clear that f then commutes with every termoperation.

9.2. Let G be the group on {0, 1, 2, 3} isomorphic to the direct oftwo copies of the group with two elements. Thus, in G, x + x = 0and x + y = z if {x, y, z} = {1, 2, 3}. Let σ = (12) and τ = (23)be permutations of {0, 1, 2, 3}. Then σ, τ ∈ Aut(G). Let A be thequasigroup on {0, 1, 2, 3} defined by x · y = σ(x) + τ(y). Then A isan Abelian quasigroup (by Corollary 5.9). Let p be a Mal’cev termfor quasigroups. (Two such are given in Exercise 2 of Chapter 2.)Clearly, r(u, v) = p(u · v, v · v, v) and s(u, v) = p(v · u, v · v, v) areboth in R(V (A)); see (9) of Chapter 9. If we take 0 ∈ A as ourzero element, then r · a = σ(a) and s · a = τ(a) for a ∈ A. Since σand τ do not commute, r and s do not commute, and so R(V (A))is not commutative. Hence, by Proposition 9.6, the ring R(V) is notcommutative for the variety V of quasigroups.

9.3. D is a division ring by Schur’s lemma. Namely if δ ∈ D thenboth the range of δ and the kernel (= {a ∈ M(β) : δ(a) = 0}) aresubmodules of M(β) as an R-module. Since M(β) is simple, it has nonontrivial submodules. Thus either δ = 0 or the kernel of δ is 0 andthe range is M(β), i.e., δ is an automorphism and so invertible. HenceD is a division ring.

Recall that a ∈ ιiM(β, zi), means that aj = zj for all j 6= i. Letm ∈ R have all entries 0 except mii = 1 (that is, mii = u). Thena ∈ ιiM(β, zi) if and only if m · a = a. If δ is an endomorphism andm · a = a then δ(a) = δ(m · a) = m · δ(a), showing ιiM(β, zi) is aD-subspace.

Suppose that a1, . . . , an ∈ M(β, zi) are such that ιia1, . . . , ιian areD-linearly independent and that b1, . . . , bn ∈M(β, zj). By the Jacob-son Density Theorem (Theorem 1.12 of Chapter 9 of Hungerford [49])there is an m ∈ R such that m · (ιiak) = ιibk. Let mji = φj(r(u, v))with r(u, v) ∈ Hij and let g(x) = r(x, z). Then g(ak) = bk, of course.

If A is finite then D is finite and hence its prime subfield has char-acteristic p for some prime p. Each block is a finite vector space overthe prime subfield and thus has order pi for some i.

9.4. To prove the first it clearly suffice to prove it in the case thatα/β ր γ/δ, i.e., γ = α ∨ δ and β = α ∧ δ. Suppose that [ϕ, α] ≤ β.Then [ϕ, γ] = [ϕ, α ∨ δ] = [ϕ, α] ∨ [ϕ, δ] ≤ β ∨ δ = δ. If [θ, γ] ≤ δ, then[θ, α] ≤ [θ, γ] ∧ α ≤ δ ∧ α = β. Hence (β : α) = (δ : γ).

For the second part we may again assume that α/β ր γ/δ. More-over we may assume that β = 0. Since [α, α] ≤ β ≤ δ, α and δ permuteby Theorem 6.2. Define a map τi : M(γ/δ, zi) → M(α, zi) as follows.


If x γ zi then, since γ = δ ◦ α, there is a y such that x δ y α zi.Moreover, since α ∧ δ = β = 0, y is unique. Let τi(x/δ) = y. Tosee that this is well defined suppose x δ x′ and that x δ y α zi andx′ δ y′ α zi. Then y δ ∧ α y

′, i.e., y = y′ and so τi is well defined. Nowlet τ : M(γ/δ)→M(α) be defined by τ(a)i = τi(ai).

To see that each τi, and hence τ , is one-to-one suppose that x,x′ γ zi and τi(x/δ) = τi(x

′/δ) = y. Then x δ y α zi and x′ δ y α zi.This implies that x δ x′ so x/δ = x′/δ. Now if y α zi then certainlyy δ y α zi and y γ zi. Hence τi(y/δ) = y, and thus τi is onto.

To see that τi, and hence τ , preserves addition let x, x′ ∈ zi/γand let y = τi(x/δ) and y′ = τi(x

′/δ). Then x δ y α zi and x′ δy′ α zi. It follows that d(x, zi, x

′) δ d(y, zi, y′) α d(zi, zi, zi) = zi. So

τi(d(x, zi, x′)) = d(y, zi, y

′) which shows that τi preserves addition.To see that τ respect multiplication by elements of R we need to

show that if r(u, v) ∈ Hij and x ∈ zi/γ then rj(r(x, z)) = r(τi(x), z).Let y = τi(x) so that xδyαzi. Then r(x, z) δ r(y, z) α r(zi, z) = zi,which implies that τj(r(x, z)) = r(y, z) = r(τi(x), z), completing theproof.

9.5. Let θ = Cg(a, b) ≤ β so that a β b. Let c = d(a, b, zi) where a,b ∈ zi/γ. Then c = d(a, b, zi) θ d(a, a, zi) = zi. Let g(x) = d(x, zi, b).Then g(zi) = b and

g(c) = d(d(a, b, zi), zi, b)

= d(d(a, b, zi), d(b, b, zi), d(b, b, b))

= d(d(a, b, b), d(b, b, b), d(zi, zi, b))

= d(a, b, b) = a

Hence Cg(a, b) = Cg(c, zi) (= θ). Let c ∈M =∑

M(β, zj) denotethe vector with all 0’s except with c in the ith coordinate. Let N = Rcbe the cyclic module generated by c. Now x αN y if and only ifx β y and if x, y ∈ zj/γ and we let v ∈ M be the vector whose ith

coordinate is d(x, y, zj) and whose other coordinates are 0, then v ∈ N.But then v = m · c for some m ∈ R. This implies that there is anr ∈ Hij such that r(c, z) = d(x, y, zj). Since r(zi, z) = zj , we have that〈d(x, y, zj), zj〉 ∈ Cg(c, zi) = Cg(a, b), i.e., αN ≤ Cg(a, b). But clearlyc αN zi; so that Cg(a, b) = αN . Hence Cg(a, b) corresponds to a cyclicsubmodule. Thus we have shown that in the isomorphism of β/0 ontoL(M) the image of a principal congruence is a cyclic submodule. Hencethe image of an n-generated congruence is an n-generated submodule.Of course, an element is compact in β/0 if and only if it is compact inCon A by upper continuity, and thus the isomorphism maps the setof compact element onto the set of finitely generated submodules.

CHAPTER 13 165

9.6. As in the proof of Proposition 9.4, if rij(u,v,y) ∈ Hji, themaps rij(u,v,y) 7→ rij(u,v, σ(y)) can be combined into a homomor-phism, which we again denote by σ, from R(V, λ, κ) onto R(V, λ). Letm = (mij) be in R(V, λ, κ) and let a be in

∑

M(β, zi). Then, byTheorem 9.7, m · a = (σm) · a. Hence, if m is in the kernel of σ,m ·a = (σm) ·a = 0 for all a in

∑

M(β, zi) and thus m is in the kernelof the action of R(V, λ, κ) on

∑

M(β, zi).

9.7. The maps ψ : M(β, zi)→M(β, zi) given by ψ(x) = d(x, zi, zi)can be combined in an obvious way to give a map ψ :

∑

M(β, zi) →∑

M(β, z′i). To see that ψi, and hence ψ, preserves addition let x,x′ ∈M(β, zi). Then

ψi(d(x, zi, x′)) = d(d(x, zi, x

′), zi, z′i)

= d(d(x, zi, x′), d(zi, zi, zi), d(z

′i, z

′i, z

′i))

= d(d(x, zi, z′i), d(zi, zi, z

′i), d(x

′, zi, z′i))

= d(ψi(x), z′i, ψi(x

′))

as desired.The proof that ψ preserves R-multiplication and is a bijection is

similar to the proof of Proposition 9.11.

Chapter 10

10.1. Let β ≺ α in Con C and let δ be a completely meet irre-ducible element with δ ≥ β and δ 6≥ α. Let γ = δ∗ = α ∨ δ be theunique cover of δ. By Exercise 4 of Chapter 9. (β : α) = (δ : γ).Let η′i =

∧

j 6=i ηj . If δ ≥ η′i then we can use induction on n to show

that (δ : γ) ≥ ηj for some j. Thus we may assume that ηi, η′i 6≤ δ.

Hence by the implication displayed in the proof of Theorem 10.1 wehave that ηi ≤ (δ : γ), as desired. The second part follows by lettingθ =

∧

β≺α(β : α) ∈ Con B. As in Theorem 10.5, θ is nilpotent and by

the first part θ =∧ni=1(θ ∨ ηi).

Chapter 13

13.1. First we will show that β + γ ⊆ β ⊕ γ. So suppose that〈x, y〉 ∈ β + γ so that there is a z and t such that x β z α2 y, x α2 t γ yand z α1 t. Then x α2 t α1 z α2 y, so 〈x, y〉 ∈ α1 ∨ α2. Sinceα1 ≤ α1 ∨ α3 = γ13 ∨ α3 and z α1 t, there is a w ∈ A such thatz γ13 w α3 t. Now we have that x β z γ13 w and

x α2 t α3 w.


so that 〈x, w〉 ∈ (β ∨ γ13) ∧ (α2 ∨ α3). Similarly 〈w, y〉 ∈ (γ ∨ α3) ∧(α2 ∨ γ13). Hence 〈x, y〉 ∈ (β ∨ γ13) ∧ (α2 ∨ α3) ∨ (γ ∨ α3) ∧ (α2 ∨ γ13),proving that 〈x, y〉 ∈ β ⊕ γ.

Next we observe that β + γ is a congruence. To see symmetrysuppose that 〈x, y〉 ∈ β + γ and that z and t are as above. Then〈y, x〉 ∈ β + γ with z′ = d(x, z, y) and t′ = d(x, t, y). Similarly β + γis transitive and it obviously respects the operations. Moreover α1 ≤α2∨ (β +γ). To see this let 〈u, v〉 ∈ α1. By hypothesis α1 ≤ β∨α2 andα1 ≤ γ ∨ α2. Hence there are elements r and s with u β r α2 v, andu α2 s γ v. By definition this implies that 〈r, s〉 ∈ β + γ. Now u α2

s β + γ r α2 v, showing that α1 ≤ α2∨ (β + γ). Since α2 ∧ (β⊕ γ) = 0,α2 ∧ (β + γ) = 0. Since in a modular lattice comparable complementsare equal, β ⊕ γ = β + γ.

13.2. First suppose that V satisfies ε3, that {αi, γij} is an n-frame,n ≥ 3, in Con C, C ∈ V and that α1 ∧ α2 = 0. Let x γ12 z. Sinceγ12 ≤ α1 ∨ α2 there is a t with x α1 t α2 z. Since α1 ≤ α2 ∨ γ12 thereis a y with x α2 y γ12 t. We have the situation indicated in Figure 1.

x z

y tγ12

α1α2

Figure 1.

Now we can apply Theorem 5.5(iii) with z for u′ and x for both yand u. By that theorem we have the relationships indicated in Figure 2.

x zd(x, z, y)

y t

γ12

α1α2

Figure 2.

Since γ12 ≤ α1 ∨ α2 there is an element t2 such that y α1 t2 α2 t.Similarly there is a y2 such that y α2 y2 γ12 t2. By Theorem 5.5(iii)d(y, t, y) and the other elements satisfy the relations indicated in Fig-ure 3.

Clearly d(x, z, x) α2 d(y, t, y). Hence we can apply Theorem 5.5(iii)again to obtain the following relationships indicated in Figure 4, where3 · x = d(x, z, d(x, z, x)).

CHAPTER 13 167

x zd(x, z, y)

yt

t2y2

d(y, t, y)

γ12

α1α2

Figure 3.

x zd(x, z, y)3 · x

yt

t2y2

d(y, t, y)

γ12

α1α2

Figure 4.

Recall the definition of γ12 + γ12 from the previous exercise. Now〈y, z〉 ∈ γ12 +γ12 by taking x and t as the two auxiliary points requiredin that definition. From this it follows that 〈z, y2〉 ∈ γ12 + γ12 + γ12by taking y and t2 as auxiliary points. Since we are assuming ε3,γ12 + γ12 + γ12 = α1. Thus y2 α1 z, and hence 3 · x α1 ∧ γ12 z; seeFigure 4. Since α1 ∧ γ12 = 0, z = 3 · x. Thus we have shown thatthe Abelian group M(γ12, z) has exponent 3 or is trivial. Now supposethat θ is an arbitrary Abelian congruence on an algebra A, and letB = A(θ) = {〈x, y〉 ∈ A2 : x θ y}. Then since θ is Abelian, the set{0, η0, η1,∆θ,θ, θ0 = θ1} is isomorphic toM3. LetC = {〈a0, . . . , an−1〉 ∈An : a0 θ a1 θ · · · θ an−1}. Then it is not hard to verify that αi =∧

j 6=i ηj, γij = ∆ij ∧ (αi ∨ αj) is an n-frame (with 0 for its index base)

in Con C, where ∆ij = {〈〈a0, . . . , an−1〉, 〈b0, . . . , bn−1〉〉 : 〈ai, aj〉 ∆θ,θ

〈bi, bj〉}. Now since θ0/η0 and γ12/0 are projective intervals, we have

MA(θ, z) ∼= MC(θ0/η0, 〈z, . . . , z〉/η0)∼= MC(γ12, 〈z, . . . , z〉)

by (the solution to) Exercise 4 of Chapter 9. Thus we have shown thatM(θ, z) has exponent 3 for every (nontrivial) Abelian congruence θ.Now the elements of R(V) lie in an Abelian congruence (in the nota-tion of Chapter 9, θ0/[θ0, θ0]) and their addition is inherited from theaddition on M(θ/[θ, θ], v) (suppressing the zeros). Thus R(V) satisfies1 + 1 + 1 = 0, as desired.


For the converse suppose that R(V) has characteristic 3 and let{αi, γij} be an n-frame in Con A, where A is in V. Let 〈z, y2〉 ∈γ12 + γ12 + γ12. Then there exist y, t2 ∈ A such that the relationsin Figure 4 hold for the four elements, z, y, y2, t2. By Lemma 13.1and Theorem 6.2, α1, α2, and γ12 pairwise permute. From this itfollows that there exists elements x and t as in Figure 4. ApplyingTheorem 5.5(iii) three times we have the full situation indicated inFigure 4.

Of course γ12/[γ12, γ12] is an Abelian congruence and so has expo-nent 3 since R(V) has characteristic 3. Thus 3 ·x [γ12, γ12] z, and since[γ12, γ12] ≤ α1 we have that 3 · z α1 z. Now we have that y2 α1 z; seeFigure 4. Thus we have shown that γ12 + γ12 + γ12 ≤ α1 which impliesthat they are equal as above. Since γ12 + γ12 + γ12 = γ12⊕ γ12⊕ γ12 bythe previous exercise, the characteristic of the ring of the frame is 3.Thus ε3 holds.

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Index

k-step nilpotent, 47k-step solvable, 47n–frame, 1154-difference term, 139

Abelian, 35, 47affine, 35algebra, 11

Abelian, 35affine, 35linear, 86quasiprimal, 109

algebraic, 13annihilates, 22

central, 47centralize, 21centralizes, 50commutator, 7, 124commutator word, 124compact, 13compactly generated, 13completely meet irreducible, 13congruence, 13

Abelian, 35fully invariant, 126

congruence identity, 59Cube Lemma, 18

Day terms, 16, 18derived operations, 12difference term, 38discriminator variety, 109distributive, 14distributive variety, 14

equation, 12equational class, 12

equivalent algebras, 12

finite type, 11Fitting congruence, 92free algebra, 14fully invariant congruence, 126

Gumm difference term, 38

identity, 12independent varieties, 106interpretation, 11interval, 13isotopic, 111

kernel, 14

locally finite, 14

Mal’cev condition, 15Mal’cev term, 18minimal variety, 109modular, 14modular variety, 14monolith, 13

neutral, 68nilpotent of class k, 47normal subloop, 44

polynomial operations, 12polynomially equivalent algebras, 12

quasiprimal, 109quotient, 13

residuation, 33

Shifting Lemma, 16similar

173

174 INDEX

θ in A to ψ in B, 92algebras, 93

simple, 13skew, 68

congruence, 68strict refinement property, 69strictly simple, 109subdirectly irreducible, 13

term, 11term condition, 10, 22term operations, 12ternary discriminator operation, 109ternary group, 35transpose, 14type, 11

variety, 12Abelian, 35

commutator theory for congruence modular varieties ralph freese

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